7.4, #9. Set t^{2} sin t = uv′, where u = t^{2}, v′ = sin t. Then u′ = 2t, v = − cos t, and

∫ t^{2} sin t dt = −t^{2} cos t + ∫ 2t cos t dt

The second integral is 2t sin t + 2 cos t + C (See example 2 on page 360 of the text). The final answer is −t^{2} cos t + 2t sin t + 2 cos t + C.

7.4, #14. Set sin^{2} θ = uv′, where u = sin θ, v′ = sin θ. Then u′ = cos θ, v = − cos θ, and

∫ sin^{2} θ dθ = − sin θ cos θ + ∫ cos^{2} θ dθ = − sin θ cos θ + ∫ (1 − sin^{2} θ) dθ

using the identity sin^{2} θ + cos^{2} θ = 1. Thus 2 ∫ sin^{2} θ dθ = − sin θ cos θ + ∫ dθ. The final

answer is (1/2)(− sin θ cos θ + θ) + C. This problem is similiar to the integral of cos^{2} θ which is done in Example 6 of page 363. Another approach is to use the trig identites:

sin^{2} θ = \frac{1}{2}(1 − cos(2θ))  and  sin(2θ) = 2 sin θ cos θ

The integral becomes

\frac{1}{2} ∫ (1 − cos(2θ)) dθ = \frac{1}{2}θ − \frac{1}{4} sin(2θ) = \frac{1}{2}(θ − sin θ cos θ).

7.4, #17. Set (ln x)/x^{2} = uv′, where u = ln x, v′ = x^{−2}. Then u′ = 1/x, v = −x^{−1}, and

∫ \frac{ln x}{x^{2}} dx = −\frac{ln x}{x} + ∫ \frac{1}{x^{2}} dx = −\frac{ln x}{x} − \frac{1}{x} + C.

7.4, #31. Set e^{x} sin x = uv′, where u = sin x, v′ = e^{x}. Then u′ = cos x, v = e^{x}, and

∫ e^{x} sin x dx = e^{x} sin x − ∫ e^{x} cos x dx.

Do the second integral in a similar way, that is, view e^{x} cos x as uv′, where u = cos x, v′ = e^{x}. Then u′ = − sin x, v = e^{x}, and

∫ e^{x} cos x dx = e^{x} cos x + ∫ e^{x} sin x dx.

Plug this into the first answer to get

∫ e^{x} sin x dx = e^{x} sin x − (e^{x} cos x + ∫ e^{x} sin x dx).

Move the integral on the right-side to the left-side and divide by 2 to get the final answer (1/2)e^{x}(sin x − cos x) + C.

7.5, #6. Use formula 12 on page 366 with x = θ, ax = 3θ, and bx = 5θ. Then

∫ sin 3θ cos 5θ dθ = \frac{1}{16}(5 sin 3θ sin 5θ + 3 cos 3θ cos 5θ) + C.

7.5, #8. Use formula 9 on page 366 with a = −3 and b = 1. Then

∫ e^{−3θ} cos θ dθ = \frac{1}{10}e^{−3θ}(−3 cos θ + sin θ) + C.

7.5, #13. Use formula 24 on page 367 with y = x and a = \sqrt{3}. Then

∫ \frac{1}{3 + y^{2}} dy = \frac{1}{\sqrt{3}} arctan \frac{y}{\sqrt{3}} + C.

7.5, #14. Use formula 14 on page 366 with p(x) = x^{2} and a = 3. Then

∫ x^{2}e^{3x} dx = \frac{1}{3}x^{2}e^{3x} − \frac{1}{9}2xe^{3x} + \frac{1}{27}2e^{3x} + C = \frac{e^{3x}}{27} (9x^{2} − 6x + 2) + C.