Use the identity \( \sin^2(3x) = \tfrac{1}{2}(1 - \cos 6x) \).

Then

\[ \int_0^{\pi/2} \sin^2(3x)\,dx = \int_0^{\pi/2} \Bigl(\tfrac{1}{2} - \tfrac{1}{2}\cos 6x\Bigr)dx = \Bigl[\tfrac{1}{2}x - \tfrac{1}{12}\sin 6x\Bigr]_0^{\pi/2}. \]

At \( x = \tfrac{\pi}{2} \): \( \tfrac{1}{2}\cdot\tfrac{\pi}{2} - \tfrac{1}{12}\sin 3\pi = \tfrac{\pi}{4} - 0 \).

At \( x = 0 \): the value is 0.

So the exact value is \( \dfrac{\pi}{4} \approx 0.785 \) (3 s.f.).

Answer: 0.785.

Use \( 1 + \tan^2(2x) = \sec^2(2x) \) and \( \dfrac{d}{dx}(\tan 2x) = 2\sec^2(2x) \).

So

\[ \int_0^{\pi/12} (1 + \tan^2(2x))\,dx = \int_0^{\pi/12} \sec^2(2x)\,dx = \Bigl[\tfrac{1}{2}\tan 2x\Bigr]_0^{\pi/12}. \]

At \( x = \tfrac{\pi}{12} \): \( 2x = \tfrac{\pi}{6} \), so \( \tan \tfrac{\pi}{6} = \tfrac{1}{\sqrt{3}} \).

Thus the value is \( \tfrac{1}{2}\cdot \tfrac{1}{\sqrt{3}} = \tfrac{1}{2\sqrt{3}} \approx 0.289 \) (3 s.f.).

Answer: 0.289.

Expand the square:

\[ (1 - \cos 2x)^2 = 1 - 2\cos 2x + \cos^2 2x. \]

Use \( \cos^2 2x = \tfrac{1}{2}(1 + \cos 4x) \):

\[ 1 - 2\cos 2x + \cos^2 2x = 1 - 2\cos 2x + \tfrac{1}{2}(1 + \cos 4x) = \tfrac{3}{2} - 2\cos 2x + \tfrac{1}{2}\cos 4x. \]

So

\[ \int_0^{\pi/2} (1 - \cos 2x)^2\,dx = \int_0^{\pi/2} \Bigl(\tfrac{3}{2} - 2\cos 2x + \tfrac{1}{2}\cos 4x\Bigr)dx = \Bigl[\tfrac{3}{2}x - \sin 2x + \tfrac{1}{8}\sin 4x\Bigr]_0^{\pi/2}. \]

At \( x = \tfrac{\pi}{2} \): \( \tfrac{3}{2}\cdot\tfrac{\pi}{2} = \tfrac{3\pi}{4} \), and \( \sin \pi = 0, \sin 2\pi = 0 \).

At \( x = 0 \): the expression is 0.

Exact value: \( \dfrac{3\pi}{4} \approx 2.36 \) (3 s.f.).

Answer: 2.36.

Expand the square:

\[ (1 + 2\sin x)^2 = 1 + 4\sin x + 4\sin^2 x. \]

Use \( \sin^2 x = \tfrac{1}{2}(1 - \cos 2x) \):

\[ 1 + 4\sin x + 4\sin^2 x = 1 + 4\sin x + 4\cdot \tfrac{1}{2}(1 - \cos 2x) = 3 + 4\sin x - 2\cos 2x. \]

So

\[ \int_0^{\pi} (1 + 2\sin x)^2\,dx = \int_0^{\pi} (3 + 4\sin x - 2\cos 2x)\,dx = \Bigl[3x - 4\cos x - \sin 2x\Bigr]_0^{\pi}. \]

At \( x = \pi \): \( 3\pi - 4\cos \pi - \sin 2\pi = 3\pi - 4(-1) - 0 = 3\pi + 4. \)

At \( x = 0 \): \( 0 - 4\cdot 1 - 0 = -4. \)

Difference: \( (3\pi + 4) - (-4) = 3\pi + 8 \approx 17.4 \) (3 s.f.).

Answer: 17.4.

Expand:

\[ (\sin x - \cos x)^2 = \sin^2 x - 2\sin x\cos x + \cos^2 x = 1 - \sin 2x, \]

since \( \sin^2 x + \cos^2 x = 1 \) and \( 2\sin x\cos x = \sin 2x \).

So

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

At \( x = \tfrac{\pi}{2} \): \( \tfrac{\pi}{2} + \tfrac{1}{2}\cos \pi = \tfrac{\pi}{2} - \tfrac{1}{2} \).

At \( x = 0 \): \( 0 + \tfrac{1}{2}\cos 0 = \tfrac{1}{2} \).

Difference: \( \tfrac{\pi}{2} - \tfrac{1}{2} - \tfrac{1}{2} = \tfrac{\pi}{2} - 1 \approx 0.571 \) (3 s.f.).

Answer: 0.571.

Use \( \displaystyle \int \sec^2 x\,dx = \tan x + C \).

\[ \int_{\pi/6}^{\pi/3} \sec^2 x\,dx = [\tan x]_{\pi/6}^{\pi/3} = \tan \tfrac{\pi}{3} - \tan \tfrac{\pi}{6} = \sqrt{3} - \tfrac{1}{\sqrt{3}}. \]

This simplifies to \( \dfrac{2}{\sqrt{3}} \approx 1.15 \) (3 s.f.).

Answer: 1.15.

Use the identity \( \cos^2 x - \sin^2 x = \cos 2x \).

Then

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

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

Answer: 0.500.

Use the product-to-sum identity:

\[ \sin A \cos B = \tfrac{1}{2}[\sin(A+B) + \sin(A-B)]. \]

With \( A = 3x \), \( B = 5x \):

\[ \sin 3x \cos 5x = \tfrac{1}{2}[\sin 8x - \sin 2x]. \]

So

\[ \int_0^{\pi/6} \sin 3x \cos 5x\,dx = \tfrac{1}{2}\int_0^{\pi/6} (\sin 8x - \sin 2x)\,dx = \Bigl[-\tfrac{1}{16}\cos 8x + \tfrac{1}{4}\cos 2x\Bigr]_0^{\pi/6}. \]

Evaluating and simplifying gives the exact value \( -\tfrac{1}{32} \approx -0.0313 \) (3 s.f.).

Answer: -0.0313.