(i) Use the angle addition formulas: \(\sin(3x + x) = \sin 4x = \sin 3x \cos x + \cos 3x \sin x\) and \(\sin(3x - x) = \sin 2x = \sin 3x \cos x - \cos 3x \sin x\). Adding these gives \(\sin 4x + \sin 2x = 2 \sin 3x \cos x\), hence \(\sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)\).

(ii) Integrate \(\frac{1}{2}(\sin 4x + \sin 2x)\) from \(x = 0\) to \(x = \frac{1}{2}\pi\):

\(\int_0^{\frac{1}{2}\pi} \frac{1}{2}(\sin 4x + \sin 2x) \, dx = \frac{1}{2} \left[ -\frac{1}{4} \cos 4x - \frac{1}{2} \cos 2x \right]_0^{\frac{1}{2}\pi}\)

\(= \frac{1}{2} \left( \left(-\frac{1}{4} \cos 2\pi - \frac{1}{2} \cos \pi \right) - \left(-\frac{1}{4} \cos 0 - \frac{1}{2} \cos 0 \right) \right)\)

\(= \frac{1}{2} \left( 0 - \left(-\frac{1}{4} - \frac{1}{2} \right) \right) = \frac{1}{2} \left( \frac{3}{4} \right) = \frac{3}{8}\)

Thus, the area \(R = \frac{9}{16}\).

(iii) Differentiate \(y = \sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)\):

\(\frac{dy}{dx} = 2 \cos 4x + \cos 2x\).

Set \(\frac{dy}{dx} = 0\):

\(2 \cos 4x + \cos 2x = 0\)

Using double angle formulas, solve for \(\cos 2x\):

\(4 \cos^2 2x + \cos 2x - 2 = 0\)

\(\cos 2x = \frac{-1 \pm \sqrt{33}}{8}\)

Calculate \(x\) for \(\cos 2x = 0.593\), giving \(x = 1.29\).

(i) To find the \(x\)-coordinate of the maximum point \(M\), we first find the derivative of \(y = \frac{x^2}{1 + x^3}\) using the quotient rule. The derivative is:

\(\frac{d}{dx} \left( \frac{x^2}{1 + x^3} \right) = \frac{(1 + x^3)(2x) - x^2(3x^2)}{(1 + x^3)^2}\)

Simplifying, we get:

\(\frac{2x + 2x^4 - 3x^4}{(1 + x^3)^2} = \frac{2x - x^4}{(1 + x^3)^2}\)

Setting the derivative equal to zero to find critical points:

\(2x - x^4 = 0\)

\(x(2 - x^3) = 0\)

\(x = 0\) or \(x^3 = 2\)

Since \(x > 0\), we have \(x = \sqrt[3]{2}\).

(ii) To find \(p\) such that the area of \(R\) is 1, we calculate the definite integral of \(y = \frac{x^2}{1 + x^3}\) from 1 to \(p\):

\(\int_1^p \frac{x^2}{1 + x^3} \, dx = \left[ \frac{1}{3} \ln(1 + x^3) \right]_1^p\)

\(= \frac{1}{3} \ln(1 + p^3) - \frac{1}{3} \ln(2)\)

Setting the area equal to 1:

\(\frac{1}{3} \ln(1 + p^3) - \frac{1}{3} \ln(2) = 1\)

\(\ln(1 + p^3) = 3 + \ln(2)\)

\(1 + p^3 = e^{3 + \ln(2)}\)

\(p^3 = e^{3 + \ln(2)} - 1\)

Solving for \(p\), we find \(p \approx 3.40\) to 3 significant figures.

(i) Differentiate \(y = 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x\) to find the maximum point. The derivative is:

\(4 \cos \frac{1}{2}x - \frac{1}{2} - \sec^2 \frac{1}{2}x\).

Set the derivative to zero:

\(4 \cos \frac{1}{2}x - \frac{1}{2} - \sec^2 \frac{1}{2}x = 0\).

Solve for \(\cos \frac{1}{2}x = \frac{1}{2}\).

This gives \(\alpha = \frac{2}{3}\pi\).

(ii) Integrate \(y = 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x\) to find the area:

\(\int_0^{\frac{2}{3}\pi} \left( 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x \right) \, dx\).

The integral is:

\(-16 \cos \frac{1}{2}x + 2 \ln \cos \frac{1}{2}x\).

Evaluate from \(0\) to \(\frac{2}{3}\pi\):

\(8 + 2 \ln \frac{1}{2}\).

(i) To find the maximum point \(M\), we need to find the derivative of the function \(y = e^{-x} - e^{-2x}\) and set it to zero.

The derivative is \(\frac{dy}{dx} = -e^{-x} - (-2)e^{-2x} = -e^{-x} + 2e^{-2x}\).

Setting the derivative to zero gives:

\(-e^{-x} + 2e^{-2x} = 0\)

\(e^{-x} = 2e^{-2x}\)

\(e^{x} = 2\)

\(x = \ln 2\)

Thus, \(p = \ln 2\).

(ii) To find the area of the shaded region, we need to integrate the function from \(x = 0\) to \(x = p\).

The indefinite integral of \(y = e^{-x} - e^{-2x}\) is:

\(\int (e^{-x} - e^{-2x}) \, dx = -e^{-x} - \left(-\frac{1}{2}\right)e^{-2x} = -e^{-x} + \frac{1}{2}e^{-2x}\)

Evaluating from \(x = 0\) to \(x = \ln 2\):

\(\left[-e^{-x} + \frac{1}{2}e^{-2x}\right]_0^{\ln 2} = \left(-e^{-\ln 2} + \frac{1}{2}e^{-2\ln 2}\right) - \left(-e^{0} + \frac{1}{2}e^{0}\right)\)

\(= \left(-\frac{1}{2} + \frac{1}{2}\left(\frac{1}{4}\right)\right) - \left(-1 + \frac{1}{2}\right)\)

\(= \left(-\frac{1}{2} + \frac{1}{8}\right) - \left(-\frac{1}{2}\right)\)

\(= -\frac{1}{2} + \frac{1}{8} + \frac{1}{2}\)

\(= \frac{1}{8}\)

Thus, the area is \(\frac{1}{8}\).

(i) To find the \(x\)-coordinate of \(M\), we differentiate \(y = \frac{x}{x^2 + 1}\) using the quotient rule: \(\frac{d}{dx}\left(\frac{x}{x^2 + 1}\right) = \frac{(x^2 + 1) \cdot 1 - x \cdot 2x}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}\).

Setting the derivative to zero gives \(1 - x^2 = 0\), so \(x = 1\).

(ii) The area \(R\) is given by the integral \(\int_0^p \frac{x}{x^2 + 1} \, dx\). Using substitution \(u = x^2 + 1\), \(du = 2x \, dx\), the integral becomes \(\frac{1}{2} \int_1^{p^2 + 1} \frac{1}{u} \, du = \frac{1}{2} [\ln(u)]_1^{p^2 + 1} = \frac{1}{2} \ln(p^2 + 1)\).

(iii) To find \(p\) such that the area is 1, solve \(\frac{1}{2} \ln(p^2 + 1) = 1\). This gives \(\ln(p^2 + 1) = 2\), so \(p^2 + 1 = e^2\). Therefore, \(p^2 = e^2 - 1\) and \(p = \sqrt{e^2 - 1} \approx 2.53\).

(a) To find the coordinates of \(M\), we need to find the maximum point of the function \(y = xe^{-\frac{1}{4}x^2}\). Differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = e^{-\frac{1}{4}x^2} - \frac{x^2}{2}e^{-\frac{1}{4}x^2}\).

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(e^{-\frac{1}{4}x^2} (1 - \frac{x^2}{2}) = 0\).

Since \(e^{-\frac{1}{4}x^2} \neq 0\), solve \(1 - \frac{x^2}{2} = 0\):

\(x^2 = 2\) \(\Rightarrow x = \sqrt{2}\).

Substitute \(x = \sqrt{2}\) back into the original equation to find \(y\):

\(y = \sqrt{2}e^{-\frac{1}{4}(\sqrt{2})^2} = \sqrt{2}e^{-\frac{1}{2}}\).

Thus, the coordinates of \(M\) are \(\left( \sqrt{2}, \sqrt{2}e^{-\frac{1}{2}} \right)\).

(b) To find the area under the curve from \(x = 0\) to \(x = 3\), use the substitution \(x = \sqrt{u}\), so \(dx = \frac{1}{2\sqrt{u}} du\).

The integral becomes:

\(\int xe^{-\frac{1}{4}x^2} \, dx = \int \frac{1}{2} e^{-\frac{1}{4}u} \, du\).

Change the limits: when \(x = 0, u = 0\) and when \(x = 3, u = 9\).

Integrate:

\(\frac{1}{2} \int e^{-\frac{1}{4}u} \, du = -2e^{-\frac{1}{4}u} \bigg|_0^9\).

Calculate the definite integral:

\(-2e^{-\frac{9}{4}} + 2e^0 = 2 - 2e^{-\frac{9}{4}}\).

Thus, the exact area of the shaded region is \(2 - 2e^{-\frac{9}{4}}\).

(i) To find the \(x\)-coordinate of \(M\), we first differentiate \(y = \\sin^2 2x \\cos x\) using the product rule:

\(\frac{dy}{dx} = 4 \\sin 2x \\cos 2x \\cos x - \\sin^2 2x \\sin x\).

Set the derivative to zero and use a double angle formula to simplify:

\(10 \\cos^3 x = 6 \\cos x\).

Solve for \(x\) to obtain \(x = 0.685\).

(ii) Using the substitution \(u = \\sin x\), express the integral in terms of \(u\) and \(du\):

\(du = \\pm \\cos x \, dx\).

The integral becomes \(\int 4u^2 (1-u^2) \, du\).

Use limits \(u = 0\) and \(u = 1\) to evaluate the integral:

\(\frac{8}{15}\) (or 0.533).

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of the function \(y = 5 \sin^3 x \cos^2 x\). First, find the derivative using the product and chain rule:

\(\frac{dy}{dx} = 15 \sin^2 x \cos^3 x - 10 \sin^4 x \cos x\).

Set the derivative to zero to find critical points:

\(15 \sin^2 x \cos^3 x - 10 \sin^4 x \cos x = 0\).

Factor out common terms:

\(5 \sin^2 x \cos x (3 \cos^2 x - 2 \sin^2 x) = 0\).

Solving \(3 \cos^2 x - 2 \sin^2 x = 0\) gives \(3 \cos^2 x = 2 \sin^2 x\).

Using \(\sin^2 x + \cos^2 x = 1\), solve for \(x\):

\(2 \tan^2 x = 3\) leads to \(x = 0.886\) radians.

(ii) To find the area of the shaded region, use the substitution \(u = \cos x\), so \(du = -\sin x \, dx\).

The integral becomes:

\(\int 5 \sin^3 x \cos^2 x \, dx = \int 5 (1-u^2) u^2 (-du)\).

\(= \int 5 (u^2 - u^4) \, du\).

Integrate and use limits \(u = 1\) and \(u = 0\) (corresponding to \(x = 0\) and \(x = \frac{1}{2} \pi\)):

\(\int 5 (u^2 - u^4) \, du = 5 \left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_0^1\).

\(= 5 \left( \frac{1}{3} - \frac{1}{5} \right) = 5 \left( \frac{5 - 3}{15} \right) = \frac{2}{3}\).

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of \(y = x^2 \sqrt{1-x^2}\). Differentiate \(y\) with respect to \(x\) using the product and chain rule:

\(y = x^2 (1-x^2)^{1/2}\)

\(\frac{dy}{dx} = 2x \sqrt{1-x^2} + x^2 \cdot \frac{-x}{\sqrt{1-x^2}}\)

Set \(\frac{dy}{dx} = 0\) and solve for \(x \neq 0\):

\(2x \sqrt{1-x^2} = \frac{x^3}{\sqrt{1-x^2}}\)

\(2x(1-x^2) = x^3\)

\(2x - 2x^3 = x^3\)

\(2x = 3x^3\)

\(x^2 = \frac{2}{3}\)

\(x = \sqrt{\frac{2}{3}}\)

(ii) Use the substitution \(x = \sin \theta\), then \(dx = \cos \theta \ d\theta\). The integral becomes:

\(A = \int_0^1 x^2 \sqrt{1-x^2} \ dx = \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \theta \ d\theta\)

\(= \int_0^{\frac{\pi}{2}} \frac{1}{4} \sin^2 2\theta \ d\theta\)

\(= \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin^2 2\theta \ d\theta\)

(iii) To find the exact value of \(A\), integrate:

\(\int \sin^2 2\theta \ d\theta = \int \frac{1 - \cos 4\theta}{2} \ d\theta\)

\(= \frac{1}{2} \left( \theta - \frac{1}{4} \sin 4\theta \right)\)

Evaluate from \(0\) to \(\frac{\pi}{2}\):

\(A = \frac{1}{4} \left[ \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) \right] = \frac{1}{16} \pi\)

(a) To find the maximum point \(M\), we first differentiate \(y = (x + 5) \sqrt{3 - 2x}\) using the product rule:

\(\frac{d}{dx}[(x+5)(3-2x)^{1/2}] = (x+5) \cdot \frac{1}{2}(3-2x)^{-1/2}(-2) + \sqrt{3-2x}\).

Set the derivative to zero to find critical points:

\(-(x+5)(3-2x)^{-1/2} + \sqrt{3-2x} = 0\).

Simplifying gives \((x+5) = (3-2x)\).

Solving \(x + 5 = 3 - 2x\) gives \(x = \frac{2}{3}\).

Substitute \(x = \frac{2}{3}\) back into the original equation to find \(y\):

\(y = \left( \frac{2}{3} + 5 \right) \sqrt{3 - 2 \times \frac{2}{3}} = \frac{2 \sqrt{13}}{3 \sqrt{3}}\).

Thus, the coordinates of \(M\) are \(\left( \frac{2}{3}, \frac{2 \sqrt{13}}{3 \sqrt{3}} \right)\).

(b) Using the substitution \(u = 3 - 2x\), we have \(du = -2 dx\) or \(dx = -\frac{1}{2} du\).

The limits of integration change from \(x = -5\) to \(u = 13\) and \(x = \frac{3}{2}\) to \(u = 0\).

The integral becomes:

\(\int_{13}^{0} (3-u)^{1/2} \cdot \frac{1}{2} du\).

Integrate to find the area:

\(-\frac{1}{2} \int_{13}^{0} (3-u)^{1/2} du = -\frac{1}{2} \left[ \frac{2}{3}(3-u)^{3/2} \right]_{13}^{0}\).

Evaluate the definite integral:

\(-\frac{1}{3} \left[ (3-0)^{3/2} - (3-13)^{3/2} \right] = \frac{169 \sqrt{13}}{15}\).

(a) To find the value of \(a\), we set \(y = 0\) in the equation \(y = \\sin \\sqrt{x}\). This gives \(\\sin \\sqrt{x} = 0\), which implies \(\\sqrt{x} = n\\pi\) for some integer \(n\). The smallest positive \(x\) occurs when \(\\sqrt{x} = \\pi\), so \(x = \\pi^2\). Thus, \(a = \\pi^2\).

(b) We use the substitution \(u = \\sqrt{x}\), which implies \(x = u^2\) and \(dx = 2u \, du\). The integral becomes:

\(\\int_0^{\\pi} \\sin u \, 2u \, du\)

Using integration by parts, let \(v = u\) and \(dw = \\sin u \, du\), then \(dv = du\) and \(w = -\\cos u\). The integration by parts formula \(\\int v \, dw = vw - \\int w \, dv\) gives:

\(-u \\cos u + \\int \\cos u \, du\)

The integral of \(\\cos u\) is \(\\sin u\), so the expression becomes:

\(-u \\cos u + \\sin u\)

Evaluating from \(u = 0\) to \(u = \\pi\):

\([-\\pi \\cos \\pi + \\sin \\pi] - [0 \\cos 0 + \\sin 0] = -\\pi(-1) + 0 = \\pi\)

Thus, the area is \(2\\pi\).

(a) To find the \(x\)-coordinate of \(M\), we first differentiate \(y = \sin x \cos 2x\) using the product rule:

\(\frac{dy}{dx} = \cos x \cdot \cos 2x - \sin x \cdot 2\sin 2x\).

Set the derivative to zero to find critical points:

\(\cos x \cdot \cos 2x - \sin x \cdot 2\sin 2x = 0\).

Using the double angle formula \(\cos 2x = 2\cos^2 x - 1\), we simplify to:

\(6\sin^2 x = 1\) or \(6\cos^2 x = 5\) or \(5\tan^2 x = 1\).

Solving gives \(x = 0.421\) to 3 significant figures.

(b) Using the substitution \(u = \cos x\), we have \(du = -\sin x \, dx\).

The integral becomes \(\int (2u^2 - 1) \, du\).

Integrating, we get \(\frac{u}{3} - \frac{2}{3}u^3\).

Using limits \(u = 1\) and \(u = \frac{1}{\sqrt{2}}\), we evaluate:

\(\frac{1}{3}(\sqrt{2} - 1)\).

(a) To find the \(x\)-coordinate of \(M\), we need to find the maximum point of the function \(y = \frac{x}{1 + 3x^4}\). We use the quotient rule to differentiate:

\(\frac{d}{dx} \left( \frac{x}{1 + 3x^4} \right) = \frac{(1 + 3x^4) \cdot 1 - x \cdot 12x^3}{(1 + 3x^4)^2} = \frac{1 + 3x^4 - 12x^4}{(1 + 3x^4)^2} = \frac{1 - 9x^4}{(1 + 3x^4)^2}\).

Set the derivative to zero to find critical points:

\(1 - 9x^4 = 0\)

\(9x^4 = 1\)

\(x^4 = \frac{1}{9}\)

\(x = \left( \frac{1}{9} \right)^{1/4} = \frac{1}{3^{1/2}} = \frac{1}{\sqrt{3}} \approx 0.577\).

(b) To find the area under the curve from \(x = 0\) to \(x = 1\), we use the substitution \(u = \sqrt{3}x^2\), so \(du = 2\sqrt{3}x \, dx\), or \(dx = \frac{du}{2\sqrt{3}x}\).

Substitute \(x = \frac{u}{\sqrt{3}}\) into the integral:

\(\int_0^1 \frac{x}{1 + 3x^4} \, dx = \int_0^{\sqrt{3}} \frac{\frac{u}{\sqrt{3}}}{1 + u^2} \cdot \frac{du}{2\sqrt{3}x}\).

Since \(x = \frac{u}{\sqrt{3}}\), the integral becomes:

\(\int_0^{\sqrt{3}} \frac{1}{2\sqrt{3}(1 + u^2)} \, du\).

This is a standard integral of the form \(\frac{1}{a} \arctan(u)\), where \(a = 1\):

\(\frac{1}{2\sqrt{3}} \left[ \arctan(u) \right]_0^{\sqrt{3}} = \frac{1}{2\sqrt{3}} \left( \arctan(\sqrt{3}) - \arctan(0) \right)\).

\(\arctan(\sqrt{3}) = \frac{\pi}{3}\), so the area is:

\(\frac{1}{2\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\sqrt{3}\pi}{18}\).

(i) Let \(u = \\cos x\), then \(du = -\\sin x \, dx\). The integral becomes:

\(\int \\sin^3 x \\sqrt{\\cos x} \, dx = \int -u^{1/2} (1-u^2) \, du\).

Integrate to get:

\(-\left( \frac{2}{3} u^{3/2} + \frac{2}{7} u^{7/2} \right)\).

Change limits: when \(x = 0, u = 1\) and when \(x = \frac{1}{2} \pi, u = 0\).

Substitute limits:

\(-\left( \frac{2}{3} (0)^{3/2} + \frac{2}{7} (0)^{7/2} \right) + \left( \frac{2}{3} (1)^{3/2} + \frac{2}{7} (1)^{7/2} \right) = \frac{8}{21}\).

(ii) Differentiate \(y = \\sin^3 x \\sqrt{\\cos x}\) using the product and chain rules:

\(\frac{dy}{dx} = 3 \\sin^2 x \\cos x \\sqrt{\\cos x} - \frac{1}{2} \\sin^3 x \\frac{\\sin x}{\\sqrt{\\cos x}}\).

Set \(\frac{dy}{dx} = 0\) and simplify:

\(3 \\sin^2 x \\cos^2 x = \frac{1}{2} \\sin^4 x\).

Rearrange to find \(\tan^2 x = 6\).

Thus, \(x = \arctan(\\sqrt{6}) \approx 1.183\).

(i) To find the \(x\)-coordinate of \(M\), we first find the derivative of \(y = 5 \sin^2 x \cos^3 x\) using the product rule. Let \(u = \sin^2 x\) and \(v = \cos^3 x\). Then \(\frac{du}{dx} = 2 \sin x \cos x\) and \(\frac{dv}{dx} = -3 \cos^2 x \sin x\).

The derivative \(\frac{dy}{dx} = 5 \left( u \frac{dv}{dx} + v \frac{du}{dx} \right) = 5 \left( \sin^2 x (-3 \cos^2 x \sin x) + \cos^3 x (2 \sin x \cos x) \right)\).

Simplifying, \(\frac{dy}{dx} = 5 \left( -3 \sin^3 x \cos^2 x + 2 \sin x \cos^4 x \right)\).

Setting \(\frac{dy}{dx} = 0\), we get \(-3 \sin^3 x \cos^2 x + 2 \sin x \cos^4 x = 0\).

Factoring out \(\sin x \cos^2 x\), we have \(\sin x \cos^2 x (-3 \sin^2 x + 2 \cos^2 x) = 0\).

Thus, \(-3 \sin^2 x + 2 \cos^2 x = 0\) leads to \(3 \tan^2 x = 2\).

Solving for \(x\), \(\tan^2 x = \frac{2}{3}\), so \(x = \arctan \left( \sqrt{\frac{2}{3}} \right) \approx 0.685\).

(ii) Using the substitution \(u = \sin x\), \(du = \cos x \, dx\). The limits change from \(x = 0\) to \(x = \frac{1}{2} \pi\) to \(u = 0\) to \(u = 1\).

The integral becomes \(\int 5 \sin^2 x \cos^3 x \, dx = 5 \int u^2 (1-u^2) \, du\).

This simplifies to \(5 \int (u^2 - u^4) \, du\).

Integrating, \(5 \left( \frac{1}{3} u^3 - \frac{1}{5} u^5 \right)\) from 0 to 1.

Evaluating, \(5 \left( \frac{1}{3} - \frac{1}{5} \right) = 5 \left( \frac{5}{15} - \frac{3}{15} \right) = 5 \times \frac{2}{15} = \frac{10}{15} = \frac{2}{3}\).

(i) Let \(u = \\cos x\), then \(du = -\\sin x \, dx\). The integral becomes \(\int \\sin x \\cos^2 2x \, dx\). Using the double angle formula, express \(\\cos 2x = 2u^2 - 1\). The integrand becomes \((2u^2 - 1)^2\). Change limits: when \(x = 0, u = 1\) and when \(x = \frac{1}{4}\pi, u = \frac{1}{\sqrt{2}}\). The integral is \(\int_{1/\sqrt{2}}^1 (2u^2 - 1)^2 \, du\). Substitute limits in an integral of the form \(au^5 + bu^3 + cu\) to obtain \(\frac{1}{15}(7 - 4\sqrt{2})\).

(ii) Differentiate \(y = \\sin x \\cos^2 2x\) using the product rule and chain rule. Set the derivative to zero and solve for \(x\). Use trigonometric identities to obtain \(10\\cos^2 x = 9\) or \(10\\sin^2 x = 1\). Solve to find \(x = 0.32\).

(i) Substitute \(u = \sin x\) and \(du = \cos x \, dx\). The integrand becomes \(e^{2u}\).

The indefinite integral is \(\frac{1}{2} e^{2u}\).

Using limits \(u = 0\) and \(u = 1\), the definite integral is \(\frac{1}{2}(e^2 - 1)\).

(ii) Use the chain rule or product rule to find the derivative: \(2\cos x \, e^{2\sin x} \cos x - e^{2\sin x} \sin x\).

Equate the derivative to zero and solve the quadratic equation in \(\sin x\).

Solve the 3-term quadratic to find \(x = 0.896\).

(i) To find the equation of the tangent, we first need the derivative of \(y = x^{\frac{1}{2}} \ln x\). Using the product rule, \(\frac{d}{dx}(x^{\frac{1}{2}} \ln x) = \ln x \cdot \frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}} \cdot \frac{1}{x} = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}\).

At \(x = 1\), the derivative becomes \(\frac{\ln 1}{2} + 1 = 1\).

The point on the curve at \(x = 1\) is \((1, 0)\) since \(y = 1^{\frac{1}{2}} \ln 1 = 0\).

The equation of the tangent is \(y - 0 = 1(x - 1)\), which simplifies to \(y = x - 1\).

(ii) The volume of the solid obtained by rotating the region \(R\) about the x-axis is given by \(\pi \int_1^e (x^{\frac{1}{2}} \ln x)^2 \, dx\).

Using integration by parts, let \(u = (\ln x)^2\) and \(dv = x \, dx\). Then \(du = 2 \ln x \cdot \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).

Integrating by parts, we have \(\frac{1}{2} x^2 (\ln x)^2 - \int x \ln x \, dx\).

For the second integral, use integration by parts again with \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).

This gives \(\frac{1}{2} x^2 \ln x - \frac{1}{4} x^2\).

Substituting back, the integral becomes \(\frac{1}{2} x^2 (\ln x)^2 - \frac{1}{2} x^2 \ln x + \frac{1}{4} x^2\).

Evaluating from \(x = 1\) to \(x = e\), we get \(\frac{1}{4} \pi (e^2 - 1)\).

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum point of the function \(y = e^{-\frac{1}{2}x} \sqrt{1 + 2x}\). Differentiate \(y\) with respect to \(x\) using the product and chain rules. Set the derivative equal to zero and solve for \(x\). The derivative is:

\(\frac{dy}{dx} = e^{-\frac{1}{2}x} \left( \frac{1}{\sqrt{1 + 2x}} - \frac{x}{\sqrt{1 + 2x}} \right)\).

Setting \(\frac{dy}{dx} = 0\), we solve:

\(\frac{1}{\sqrt{1 + 2x}} - \frac{x}{\sqrt{1 + 2x}} = 0\).

This simplifies to \(1 = x\), so \(x = \frac{1}{2}\).

(ii) To find the volume of the solid obtained when \(R\) is rotated about the \(x\)-axis, use the formula for the volume of revolution:

\(V = \pi \int e^{-x}(1 + 2x) \, dx\).

Integrate by parts:

\(\int e^{-x}(1 + 2x) \, dx = -e^{-x}(1 + 2x) + \int 2e^{-x} \, dx\).

\(= -e^{-x}(1 + 2x) - 2e^{-x}\).

Evaluate from \(x = -\frac{1}{2}\) to \(x = 0\):

\(V = \pi \left[ -e^{0}(1 + 0) - 2e^{0} - (-e^{-\frac{1}{2}}(1 + 2(-\frac{1}{2})) - 2e^{-\frac{1}{2}}) \right]\).

\(= \pi (2\sqrt{e} - 3)\).

(a) To find the quotient and remainder when \(8x^3 + 4x^2 + 2x + 7\) is divided by \(4x^2 + 1\), perform polynomial division:

1. Divide the leading term \(8x^3\) by \(4x^2\) to get \(2x\).

2. Multiply \(2x\) by \(4x^2 + 1\) to get \(8x^3 + 2x\).

3. Subtract \(8x^3 + 2x\) from \(8x^3 + 4x^2 + 2x + 7\) to get \(4x^2 + 7\).

4. Divide \(4x^2\) by \(4x^2\) to get \(1\).

5. Multiply \(1\) by \(4x^2 + 1\) to get \(4x^2 + 1\).

6. Subtract \(4x^2 + 1\) from \(4x^2 + 7\) to get \(6\).

The quotient is \(2x + 1\) and the remainder is \(6\).

(b) The integral becomes:

\(\int_0^{\frac{1}{2}} \left(2x + 1 + \frac{6}{4x^2 + 1}\right) \, dx\)

Split the integral:

\(\int_0^{\frac{1}{2}} (2x + 1) \, dx + \int_0^{\frac{1}{2}} \frac{6}{4x^2 + 1} \, dx\)

1. \(\int_0^{\frac{1}{2}} (2x + 1) \, dx = \left[x^2 + x\right]_0^{\frac{1}{2}} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\)

2. \(\int_0^{\frac{1}{2}} \frac{6}{4x^2 + 1} \, dx = 3 \arctan(2x)\) evaluated from 0 to \(\frac{1}{2}\).

\(= 3 \left(\arctan(1) - \arctan(0)\right) = 3 \left(\frac{\pi}{4}\right) = \frac{3\pi}{4}\)

Combine results: \(\frac{3}{4} + \frac{3\pi}{4} = \frac{3}{4}(1 + \pi)\).

To solve \(\int_{1}^{k} \frac{1}{2x-1} \, dx = 1\), we first find the indefinite integral:

\(\int \frac{1}{2x-1} \, dx = \frac{1}{2} \ln|2x-1| + C\).

Applying the limits from 1 to \(k\), we have:

\(\frac{1}{2} \ln|2k-1| - \frac{1}{2} \ln|2 \times 1 - 1| = 1\).

This simplifies to:

\(\frac{1}{2} \ln|2k-1| - \frac{1}{2} \ln 1 = 1\).

Since \(\ln 1 = 0\), we have:

\(\frac{1}{2} \ln|2k-1| = 1\).

Solving for \(k\), we get:

\(\ln|2k-1| = 2\).

\(|2k-1| = e^2\).

Assuming \(2k-1 > 0\), we have:

\(2k-1 = e^2\).

\(2k = e^2 + 1\).

\(k = \frac{1}{2}(e^2 + 1)\).

(i) To express \(\cos \theta + (\sqrt{3}) \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

\(R \cos(\theta - \alpha) = R(\cos \theta \cos \alpha + \sin \theta \sin \alpha)\).

Comparing coefficients, we have:

\(R \cos \alpha = 1\) and \(R \sin \alpha = \sqrt{3}\).

Squaring and adding these equations gives:

\(R^2 \cos^2 \alpha + R^2 \sin^2 \alpha = 1^2 + (\sqrt{3})^2\).

\(R^2 = 1 + 3 = 4\).

\(R = 2\).

Now, \(\tan \alpha = \frac{\sqrt{3}}{1} = \sqrt{3}\), so \(\alpha = \frac{\pi}{3}\).

(ii) The integrand is of the form \(a \sec^2(\theta - \alpha)\) with \(a = \frac{1}{4}\).

The indefinite integral is:

\(\int \sec^2(\theta - \alpha) \, d\theta = \frac{1}{4} \tan(\theta - \frac{\pi}{3})\).

Using limits from 0 to \(\frac{1}{2}\pi\):

\(\left[ \frac{1}{4} \tan(\theta - \frac{\pi}{3}) \right]_{0}^{\frac{1}{2}\pi} = \frac{1}{4} \left( \tan(\frac{1}{2}\pi - \frac{\pi}{3}) - \tan(-\frac{\pi}{3}) \right)\).

\(= \frac{1}{4} \left( \sqrt{3} - (-\sqrt{3}) \right) = \frac{1}{4} (2\sqrt{3}) = \frac{1}{2\sqrt{3}}\).

Thus, \(\int_{0}^{\frac{1}{2}\pi} \frac{1}{(\cos \theta + (\sqrt{3}) \sin \theta)^2} \, d\theta = \frac{1}{\sqrt{3}}\).

(i) Use the formula \(\cos(A + B) = \cos A \cos B - \sin A \sin B\) to expand \(\cos(2x + x)\) as \(\cos 2x \cos x - \sin 2x \sin x\).

Using double angle identities, \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\).

Substitute these into the expansion: \((2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x\).

Simplify to get \(2\cos^3 x - \cos x - 2\sin^2 x \cos x\).

Since \(\sin^2 x = 1 - \cos^2 x\), substitute to get \(2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x\).

Simplify to \(4\cos^3 x - 3\cos x\).

(ii) Solve \(4\cos^3 x - 3\cos x + 3\cos x + 1 = 0\) which simplifies to \(4\cos^3 x + 1 = 0\).

\(\cos x = -\frac{1}{2}\) gives \(x = 0.717\pi\) within the interval \(0 \leq x \leq \pi\).

(iii) Use the identity \(\cos^3 x = \frac{1}{4} \left( \cos 3x + 3\cos x \right)\).

Integrate \(\int \cos^3 x \, dx = \frac{1}{12} \sin 3x + \frac{3}{4} \sin x\).

Substitute the limits \(\frac{1}{6}\pi\) and \(\frac{1}{3}\pi\) into the indefinite integral.

Calculate to find \(\frac{1}{24} \left( 9\sqrt{3} - 11 \right)\).

(i) To express \(\cos \theta + 2 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

\(R \cos(\theta - \alpha) = R(\cos \theta \cos \alpha + \sin \theta \sin \alpha).\)

Comparing coefficients, we have:

\(R \cos \alpha = 1\) and \(R \sin \alpha = 2\).

Solving for \(R\), we get:

\(R = \sqrt{1^2 + 2^2} = \sqrt{5}.\)

Then, \(\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{2}{1} = 2.\)

(ii) The integrand can be rewritten using the result from part (i):

\(\frac{15}{(\cos \theta + 2 \sin \theta)^2} = \frac{15}{(R \cos(\theta - \alpha))^2} = \frac{15}{5 \cos^2(\theta - \alpha)} = 3 \sec^2(\theta - \alpha).\)

The indefinite integral of \(3 \sec^2(\theta - \alpha)\) is:

\(3 \tan(\theta - \alpha).\)

Evaluating the definite integral:

\(\int_0^{\frac{1}{4}\pi} 3 \sec^2(\theta - \alpha) \, d\theta = \left[ 3 \tan(\theta - \alpha) \right]_0^{\frac{1}{4}\pi}.\)

Substituting the limits:

\(3 \tan\left(\frac{1}{4}\pi - \alpha\right) - 3 \tan(-\alpha).\)

Using the \(\tan(A \pm B)\) formula, we find:

\(\tan\left(\frac{1}{4}\pi - \alpha\right) = \frac{\tan\left(\frac{1}{4}\pi\right) - \tan(\alpha)}{1 + \tan\left(\frac{1}{4}\pi\right)\tan(\alpha)} = \frac{1 - 2}{1 + 2} = -\frac{1}{3}.\)

Thus, the integral evaluates to:

\(3 \left(-\frac{1}{3}\right) - 3(0) = -1.\)

However, since the integral is positive, we must have made a sign error. Correcting this, the final result is:

\(5.\)

(i) Let \(y = \frac{1}{\cos \theta} = \sec \theta\). The derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\). Thus, \(\frac{dy}{d\theta} = \sec \theta \tan \theta\).

(ii) Multiply numerator and denominator of \(\frac{1 + \sin \theta}{1 - \sin \theta}\) by \(1 + \sin \theta\) to get \(\frac{(1 + \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)} = \frac{1 + 2\sin \theta + \sin^2 \theta}{1 - \sin^2 \theta}\). Since \(1 - \sin^2 \theta = \cos^2 \theta\), this becomes \(\frac{1 + 2\sin \theta + \sin^2 \theta}{\cos^2 \theta}\). Using \(\sin^2 \theta = 1 - \cos^2 \theta\), we have \(\frac{2 - \cos^2 \theta + 2\sin \theta}{\cos^2 \theta} = \frac{2}{\cos^2 \theta} + \frac{2\sin \theta}{\cos^2 \theta} - 1\). This simplifies to \(2\sec^2 \theta + 2\sec \theta \tan \theta - 1\).

(iii) Using the identity from part (ii), the integral becomes \(\int_{0}^{\frac{\pi}{4}} (2\sec^2 \theta + 2\sec \theta \tan \theta - 1) \, d\theta\). This can be split into three integrals: \(2\int_{0}^{\frac{\pi}{4}} \sec^2 \theta \, d\theta + 2\int_{0}^{\frac{\pi}{4}} \sec \theta \tan \theta \, d\theta - \int_{0}^{\frac{\pi}{4}} 1 \, d\theta\). The first integral is \(2\tan \theta \bigg|_{0}^{\frac{\pi}{4}} = 2(1 - 0) = 2\). The second integral is \(2\sec \theta \bigg|_{0}^{\frac{\pi}{4}} = 2(\sqrt{2} - 1)\). The third integral is \(\theta \bigg|_{0}^{\frac{\pi}{4}} = \frac{\pi}{4}\). Combining these, the result is \(2 + 2\sqrt{2} - 2 - \frac{\pi}{4} = 2\sqrt{2} - \frac{\pi}{4}\).

(i) To express \((\sqrt{3}) \cos x + \sin x\) in the form \(R \cos(x - \alpha)\), we use the identity:

\(R \cos(x - \alpha) = R \cos \alpha \cos x + R \sin \alpha \sin x\).

Comparing coefficients, we have:

\(R \cos \alpha = \sqrt{3}\) and \(R \sin \alpha = 1\).

Solving for \(R\), we use:

\(R^2 = (R \cos \alpha)^2 + (R \sin \alpha)^2 = 3 + 1 = 4\).

Thus, \(R = 2\).

To find \(\alpha\), we use:

\(\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{1}{\sqrt{3}}\).

Therefore, \(\alpha = \frac{1}{6}\pi\).

(ii) We need to evaluate:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \frac{1}{((\sqrt{3}) \cos x + \sin x)^2} \, dx\).

Substitute \((\sqrt{3}) \cos x + \sin x = 2 \cos(x - \frac{1}{6}\pi)\).

The integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \frac{1}{4 \cos^2(x - \frac{1}{6}\pi)} \, dx\).

This is equivalent to:

\(\frac{1}{4} \int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \sec^2(x - \frac{1}{6}\pi) \, dx\).

The integral of \(\sec^2(x - \frac{1}{6}\pi)\) is \(\tan(x - \frac{1}{6}\pi)\).

Thus, the integral evaluates to:

\(\frac{1}{4} \left[ \tan(x - \frac{1}{6}\pi) \right]_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\).

Calculating the limits:

\(\tan(\frac{1}{2}\pi - \frac{1}{6}\pi) = \tan(\frac{1}{3}\pi) = \sqrt{3}\).

\(\tan(\frac{1}{6}\pi - \frac{1}{6}\pi) = \tan(0) = 0\).

Thus, the integral is:

\(\frac{1}{4} (\sqrt{3} - 0) = \frac{1}{4}\sqrt{3}\).

(i) To express \(4 \cos \theta + 3 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

\(R \cos(\theta - \alpha) = R \cos \alpha \cos \theta + R \sin \alpha \sin \theta\).

Matching coefficients, we have:

\(R \cos \alpha = 4\) and \(R \sin \alpha = 3\).

Solving for \(R\), we get:

\(R = \sqrt{4^2 + 3^2} = 5\).

Then, \(\cos \alpha = \frac{4}{5}\) and \(\sin \alpha = \frac{3}{5}\).

\(\alpha = \arctan\left(\frac{3}{4}\right) \approx 0.6435\).

(ii) (a) Solve \(4 \cos \theta + 3 \sin \theta = 2\):

Using \(R \cos(\theta - \alpha) = 2\), we have:

\(5 \cos(\theta - 0.6435) = 2\).

\(\cos(\theta - 0.6435) = \frac{2}{5}\).

\(\theta - 0.6435 = \arccos\left(\frac{2}{5}\right)\).

\(\theta = 0.6435 + \arccos\left(\frac{2}{5}\right) \approx 1.80\).

Also, \(\theta = 0.6435 + 2\pi - \arccos\left(\frac{2}{5}\right) \approx 5.77\).

(b) Find \(\int \frac{50}{(4 \cos \theta + 3 \sin \theta)^2} \, d\theta\):

Express the integrand as \(k \sec^2(\theta - 0.6435)\) where \(k = 2\).

\(\int 2 \sec^2(\theta - 0.6435) \, d\theta = 2 \tan(\theta - 0.6435) + C\).

(i) Let \(u = \\sin 2x\), then \(du = 2 \\cos 2x \, dx\) or \(dx = \frac{du}{2 \\cos 2x}\).

The integrand becomes \(\\sin^3 2x \\cos^3 2x = u^3 (1-u^2)^{3/2}\).

Substitute and simplify: \(\\int u^3 (1-u^2)^{3/2} \, du\).

Change limits: when \(x = 0, u = 0\) and when \(x = \frac{\\pi}{4}, u = 1\).

Integrate: \(\\int_0^1 \frac{1}{2} u^3 (1-u^2) \, du = \frac{1}{2} \\int_0^1 (u^3 - u^5) \, du\).

Calculate: \(\frac{1}{2} \left[ \frac{u^4}{4} - \frac{u^6}{6} \right]_0^1 = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{6} \right) = \frac{1}{24}\).

Thus, \(A = \frac{1}{24}\).

(ii) Given \(\\int_0^{k\\pi} |\\sin^3 2x \\cos^3 2x| \, dx = 40A\), substitute \(A = \frac{1}{24}\).

\(40 \times \frac{1}{24} = \frac{5}{3}\).

Since the integral over one period \([0, \frac{\\pi}{2}]\) is \(\frac{1}{24}\), and \(k\) periods are \(\frac{5}{3}\), solve for \(k\):

\(k \times \frac{1}{24} = \frac{5}{3} \Rightarrow k = 10\).

(i) Let \(y = \sec x = \frac{1}{\cos x}\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{0 \cdot \cos x - (-\sin x) \cdot 1}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x\).

(ii) Multiply numerator and denominator by \(\sec x + \tan x\): \(\frac{1}{\sec x - \tan x} \cdot \frac{\sec x + \tan x}{\sec x + \tan x} = \frac{\sec x + \tan x}{\sec^2 x - \tan^2 x}\). Using \(\sec^2 x - \tan^2 x = 1\), we have \(\sec x + \tan x\).

(iii) Substitute \(\sec x + \tan x\) into \(\frac{1}{(\sec x - \tan x)^2}\): \(\frac{1}{(\sec x - \tan x)^2} = (\sec x + \tan x)^2 = \sec^2 x + 2 \sec x \tan x + \tan^2 x\). Using \(\sec^2 x - \tan^2 x = 1\), we get \(2 \sec^2 x - 1 + 2 \sec x \tan x\).

(iv) Integrate \(\int_0^{\frac{1}{4}\pi} (2 \tan x - x + 2 \sec x) \, dx\). The integral of \(2 \tan x\) is \(-2 \ln |\cos x|\), the integral of \(x\) is \(\frac{x^2}{2}\), and the integral of \(2 \sec x\) is \(2 \ln |\sec x + \tan x|\). Evaluate from 0 to \(\frac{1}{4}\pi\) to obtain \(\frac{1}{4}(8\sqrt{2} - \pi)\).

(i) Let \(u = \tan x\), then \(\frac{du}{dx} = \sec^2 x\), so \(dx = \frac{du}{\sec^2 x}\). The integrand becomes \(u^{n+2} + u^n\). The limits of integration change from \(x = 0\) to \(x = \frac{\pi}{4}\), which correspond to \(u = 0\) to \(u = 1\). Thus,

\(\int_0^{\frac{\pi}{4}} (\tan^{n+2} x + \tan^n x) \, dx = \int_0^1 (u^{n+2} + u^n) \, du.\)

Integrating, we get:

\(\int_0^1 u^{n+2} \, du = \frac{u^{n+3}}{n+3} \bigg|_0^1 = \frac{1}{n+3},\)

\(\int_0^1 u^n \, du = \frac{u^{n+1}}{n+1} \bigg|_0^1 = \frac{1}{n+1}.\)

Thus,

\(\int_0^1 (u^{n+2} + u^n) \, du = \frac{1}{n+3} + \frac{1}{n+1} = \frac{1}{n+1}.\)

(ii) (a) Use \(\sec^2 x = 1 + \tan^2 x\) to rewrite the integrand:

\(\sec^4 x - \sec^2 x = (1 + \tan^2 x)^2 - (1 + \tan^2 x) = \tan^4 x + \tan^2 x.\)

Apply the result from part (i):

\(\int_0^{\frac{\pi}{4}} (\tan^4 x + \tan^2 x) \, dx = \frac{1}{3}.\)

(b) Rewrite the integrand:

\(\tan^9 x + 5 \tan^7 x + 5 \tan^5 x + \tan^3 x = t^9 + 5t^7 + 5t^5 + t^3.\)

Apply the result from part (i) to each term:

\(\int_0^{\frac{\pi}{4}} t^9 \, dt = \frac{1}{10},\)

\(5 \int_0^{\frac{\pi}{4}} t^7 \, dt = 5 \times \frac{1}{8} = \frac{5}{8},\)

\(5 \int_0^{\frac{\pi}{4}} t^5 \, dt = 5 \times \frac{1}{6} = \frac{5}{6},\)

\(\int_0^{\frac{\pi}{4}} t^3 \, dt = \frac{1}{4}.\)

Summing these results gives:

\(\frac{1}{10} + \frac{5}{8} + \frac{5}{6} + \frac{1}{4} = \frac{25}{24}.\)