Trig Integral Practice Problems with Answers and Solutions

These trig integral practice problems cover powers of sine and cosine, tangent and secant, and products of different angles. Try each one first. The answers come first so you can check quickly, then the full solutions.

The identities you will use:

Problems

Even powers: half-angle

1. \displaystyle\int\sin^2 x\,dx
2. \displaystyle\int\cos^2 x\,dx
3. \displaystyle\int\sin^2 x\cos^2 x\,dx

Odd powers: save a factor

4. \displaystyle\int\sin^2 x\cos x\,dx
5. \displaystyle\int\cos^3 x\,dx
6. \displaystyle\int\sin^3 x\,dx
7. \displaystyle\int\sin^3 x\cos^2 x\,dx

Tangent and secant

8. \displaystyle\int\tan^2 x\,dx
9. \displaystyle\int\sec^4 x\,dx
10. \displaystyle\int\tan^3 x\sec x\,dx

Different angles and definite integrals

11. \displaystyle\int\sin(3x)\cos x\,dx
12. \displaystyle\int_0^{\pi}\sin^2 x\,dx

Answers

Step-by-step solutions

1. Use the half-angle identity:

See the integral of sin² x for more.

2. Same idea with a plus sign:

3. Use \sin x\cos x = \frac{1}{2}\sin 2x, so \sin^2 x\cos^2 x = \frac{1}{4}\sin^2 2x. Then apply the half-angle identity with angle 2x:

4. Let u = \sin x, so du = \cos x\,dx:

5. Save one \cos x and write \cos^2 x = 1 - \sin^2 x. With u = \sin x:

6. Save one \sin x and write \sin^2 x = 1 - \cos^2 x. With u = \cos x, du = -\sin x\,dx:

7. Sine has the odd power, so save one \sin x: \sin^3 x\cos^2 x = (1 - \cos^2 x)\cos^2 x\sin x. With u = \cos x:

8. Use \tan^2 x = \sec^2 x - 1:

9. Save \sec^2 x for du and write the other \sec^2 x = 1 + \tan^2 x. With u = \tan x:

10. Save \sec x\tan x for du with u = \sec x. The leftover \tan^2 x = \sec^2 x - 1 = u^2 - 1:

11. Use the product-to-sum identity \sin A\cos B = \frac{1}{2}\big[\sin(A + B) + \sin(A - B)\big]:

12. From problem 1:

This makes sense: on [0, \pi], \sin^2 x averages \frac{1}{2}, and \frac{1}{2}\cdot\pi = \frac{\pi}{2}.

Common mistakes on these problems

• Using u-substitution on an even power. \int\sin^2 x\,dx is not \frac{\sin^3 x}{3}. There is no \cos x factor to make du.
• Mixing up the half-angle signs. \sin^2 gets the minus sign, \cos^2 gets the plus sign. Check with x = 0: \sin^2 0 = 0 = \frac{1 - 1}{2}.
• Forgetting the chain rule factor. \int\cos 2x\,dx = \frac{\sin 2x}{2}, not \sin 2x.
• Not saving a factor. In problem 6, all three sines must be used: one for du, two converted to cosine.

Keep practicing

For unlimited problems with instant feedback, try the trig integrals lesson. See also the integral of tan² x and the integral of sec³ x. When you see \sqrt{a^2 - x^2} or x^2 + a^2, move on to trig substitution problems.