Use \( \displaystyle \int \cos(2x)\,dx = \tfrac{1}{2}\sin(2x) + C \).

\[ \int_0^{\pi/4} \cos(2x)\,dx = \left[\tfrac{1}{2}\sin(2x)\right]_0^{\pi/4} = \tfrac{1}{2}\sin\tfrac{\pi}{2} - \tfrac{1}{2}\sin 0 = \tfrac{1}{2} - 0 = \tfrac{1}{2}. \]

As a decimal, \( \tfrac{1}{2} = 0.500 \) (3 s.f.).

Answer: 0.500.

\[ \int \sin(2x + \tfrac{\pi}{4})\,dx = -\tfrac{1}{2}\cos(2x + \tfrac{\pi}{4}) + C. \]

So

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

At \( x = \tfrac{\pi}{2} \): \( 2x + \tfrac{\pi}{4} = \tfrac{5\pi}{4} \), \( \cos \tfrac{5\pi}{4} = -\tfrac{\sqrt{2}}{2} \), so the value is \( -\tfrac{1}{2}\cdot -\tfrac{\sqrt{2}}{2} = \tfrac{1}{2\sqrt{2}} \).

At \( x = 0 \): \( 2x + \tfrac{\pi}{4} = \tfrac{\pi}{4} \), giving \( -\tfrac{1}{2}\cdot \tfrac{\sqrt{2}}{2} = -\tfrac{1}{2\sqrt{2}} \).

Difference: \( \tfrac{1}{2\sqrt{2}} - (-\tfrac{1}{2\sqrt{2}}) = \tfrac{1}{\sqrt{2}} \).

As a decimal, \( \tfrac{1}{\sqrt{2}} \approx 0.707 \) (3 s.f.).

Answer: 0.707.

Use \( \displaystyle \int \sec^2(u)\,du = \tan(u) + C \).

Let \( u = 3x + \tfrac{\pi}{12} \), then \( du = 3\,dx \), so \( dx = \tfrac{1}{3}du \).

\[ \int_0^{\pi/12} \sec^2(3x + \tfrac{\pi}{12})\,dx = \tfrac{1}{3}\int_{u(0)}^{u(\pi/12)} \sec^2(u)\,du = \tfrac{1}{3}[\tan u]_{\pi/12}^{\pi/3} = \tfrac{1}{3}\bigl(\tan \tfrac{\pi}{3} - \tan \tfrac{\pi}{12}\bigr). \]

Numerically, this is approximately \( 0.488 \) (3 s.f.).

Answer: 0.488.

\[ \int_0^{\pi/2} \cos(2x)\,dx = \left[\tfrac{1}{2}\sin(2x)\right]_0^{\pi/2} = \tfrac{1}{2}\sin \pi - \tfrac{1}{2}\sin 0 = 0 - 0 = 0. \]

Answer: 0.

Let \( u = \tfrac{3\pi}{2} - \tfrac{x}{3} \), so \( du = -\tfrac{1}{3}\,dx \) and \( dx = -3\,du \).

Then

\[ \int_0^{\pi} \cos\Bigl(\tfrac{3\pi}{2} - \tfrac{x}{3}\Bigr)\,dx = -3\int_{u(0)}^{u(\pi)} \cos(u)\,du = -3[\sin u]_{3\pi/2}^{7\pi/6} = -3\bigl(\sin \tfrac{7\pi}{6} - \sin \tfrac{3\pi}{2}\bigr). \]

\( \sin \tfrac{7\pi}{6} = -\tfrac{1}{2} \), \( \sin \tfrac{3\pi}{2} = -1 \), so

\[ -3\bigl(-\tfrac{1}{2} - (-1)\bigr) = -3\Bigl(\tfrac{1}{2}\Bigr) = -\tfrac{3}{2} \approx -1.50 \text{ (3 s.f.).} \]

Answer: -1.50.

\[ \int \sin(3x - \tfrac{\pi}{4})\,dx = -\tfrac{1}{3}\cos(3x - \tfrac{\pi}{4}) + C. \]

So

\[ \int_0^{\pi/2} \sin(3x - \tfrac{\pi}{4})\,dx = \left[-\tfrac{1}{3}\cos(3x - \tfrac{\pi}{4})\right]_0^{\pi/2}. \]

At \( x = \tfrac{\pi}{2} \): angle \( = \tfrac{5\pi}{4} \), \( \cos \tfrac{5\pi}{4} = -\tfrac{\sqrt{2}}{2} \), giving \( \tfrac{1}{3\sqrt{2}} \).

At \( x = 0 \): angle \( = -\tfrac{\pi}{4} \), \( \cos(-\tfrac{\pi}{4}) = \tfrac{\sqrt{2}}{2} \), giving \( -\tfrac{1}{3\sqrt{2}} \).

Difference: \( \tfrac{1}{3\sqrt{2}} - (-\tfrac{1}{3\sqrt{2}}) = \tfrac{\sqrt{2}}{3} \approx 0.471 \) (3 s.f.).

Answer: 0.471.

Let \( u = \tfrac{\pi}{12} - 3x \), so \( du = -3\,dx \) and \( dx = -\tfrac{1}{3}du \).

Then

\[ \int_{\pi/12}^{\pi/3} \sec^2\bigl(\tfrac{\pi}{12} - 3x\bigr)\,dx = -\tfrac{1}{3}\int_{u(\pi/12)}^{u(\pi/3)} \sec^2(u)\,du = -\tfrac{1}{3}[\tan u]_{\pi/12}^{-\pi/6}. \]

This equals \( -\tfrac{1}{3}\bigl(\tan(-\tfrac{\pi}{6}) - \tan \tfrac{\pi}{12}\bigr) \).

Using exact values, this simplifies to \( \dfrac{4}{3\sqrt{3}} \approx 0.770 \) (3 s.f.).

Answer: 0.770.

\[ \int -\tfrac{1}{3}\cos\Bigl(\tfrac{x}{2}\Bigr)\,dx = -\tfrac{2}{3}\sin\Bigl(\tfrac{x}{2}\Bigr) + C. \]

So

\[ \int_{-\pi/4}^{\pi/4} -\tfrac{1}{3}\cos\Bigl(\tfrac{x}{2}\Bigr)\,dx = \left[-\tfrac{2}{3}\sin\Bigl(\tfrac{x}{2}\Bigr)\right]_{-\pi/4}^{\pi/4} = -\tfrac{2}{3}\sin\tfrac{\pi}{8} - \Bigl(-\tfrac{2}{3}\sin(-\tfrac{\pi}{8})\Bigr) = -\tfrac{4}{3}\sin\tfrac{\pi}{8}. \]

Numerically, this is approximately \( -0.510 \) (3 s.f.).

Answer: -0.510.

Let \( u = 2 - \tfrac{x}{2} \), so \( du = -\tfrac{1}{2}\,dx \) and \( dx = -2\,du \).

When \( x = 1 \), \( u = 2 - \tfrac{1}{2} = \tfrac{3}{2} \); when \( x = 3 \), \( u = 2 - \tfrac{3}{2} = \tfrac{1}{2} \).

\[ \int_1^3 \cos\Bigl(2 - \tfrac{x}{2}\Bigr)\,dx = -2\int_{3/2}^{1/2} \cos u\,du = -2[\sin u]_{3/2}^{1/2} = -2(\sin \tfrac{1}{2} - \sin \tfrac{3}{2}). \]

Numerically this is approximately \( 1.04 \) (3 s.f.).

Answer: 1.04.