7.5, #17. Formula 15 with x = y, p(x) = y^{2}, a = 2 gives

∫ y^{2} sin 2y dy = −\frac{1}{2}y^{2} cos 2y + \frac{1}{4}2y sin 2y + \frac{1}{8}2 cos 2y + C.

7.5, #18. Formula 11 with x = y, a = 2, b = 7 gives

∫ cos 2y cos 7y dy = \frac{1}{45}(7 cos 2y sin 7y − 2 sin 2y cos 7y) + C.

7.5, #24. Substitute x = 3y, dx = 3dy to get ∫ cos^{4} 3y dy = (1/3) ∫ cos^{4} x dx. Then use formula 18 with n = 4 to get

∫ cos^{4} x dx = \frac{1}{4} cos^{3} x sin x + \frac{3}{4} ∫ cos^{2} x dx

Use formula 18 again with n = 2 to get

∫ cos^{2} x dx = \frac{1}{2} cos x sin x + ∫ dx = \frac{1}{2} cos x sin x + \frac{1}{2}x + C.

Putting these pieces together gives

∫ cos^{4} 3y dy = \frac{1}{12} cos^{3} 3y sin 3y + \frac{1}{8} cos 3y sin 3y + \frac{3}{8}y + C.

7.5, #28. Make the Substitution w = 2x, dw = 2dx and then use formula 24 to get

∫ \frac{dx}{1 + 4x^{2}} = \frac{1}{2} ∫ \frac{dw}{1 + w^{2}} = \frac{1}{2} arctan w + C = \frac{1}{2} arctan 2x + C.

Comment: A very common mistake — using formula 24 without the first making the substitution w = 2x. This gives incorrect answer arctan 2x + C.

7.5, #37. Formula 26 with x = z, a = 3, and b = 0 gives

∫ \frac{dz}{z(z − 3)} = \frac{1}{3}(ln |z − 3| − ln |z|) + C.

7.5, #38. Formula 26 with x = y, a = 2, and b = −2 gives

∫ \frac{dy}{4 − y^{2}} dy = − ∫ \frac{dy}{(y − 2)(y + 2)} = −\frac{1}{4}(ln |y − 2| − ln |y + 2|) + C.

7.5, #44. Since y^{2} + 4y + 5 = (y + 2)^{2} + 1^{2}, we substitute w = y + 2 and dw = dy to get

∫ \frac{dy}{y^{2} + 4y + 5} = ∫ \frac{dw}{w^{2} + 1^{2}} = arctan w + C = arctan(y + 2) + C.

7.5, #45. The integrand is an improper rational function, so we use long division to write

\frac{x^{3} + 3}{x^{2} − 3x + 2} = x + 3 + \frac{7x − 3}{x^{2} − 3x + 2}.

Now x^{2} − 3x + 2 = (x − 1)(x − 2), so formula 27 with a = 1, b = 2, c = 7, and d = −3 applies, giving

∫ \frac{x^{3} + 3}{x^{2} − 3x + 2} dx = \frac{1}{2}x^{2} + 3x − 4 ln |x − 1| + 11 ln |x − 2| + C.