# Trig Integral Practice Problems with Answers and Solutions

> 12 trigonometric integral practice problems with answers: powers of sine and cosine, tangent and secant, half-angle and product-to-sum.

Source: https://practicecalculus.com/practice-problems/trig-integrals
Updated: 2026-11-06

$$ \sin^2 x = \frac{1 - \cos 2x}{2}, \quad \cos^2 x = \frac{1 + \cos 2x}{2} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 2 and AP Calculus BC
- **You need:** [U-substitution](/practice-problems/u-substitution) and the Pythagorean identities
- **Main idea:** Odd power: save one factor for $du$. Even powers: use the half-angle identities
- **Practice with new numbers:** [Trig integrals lesson](/learn/calculus-2/integration-techniques/trig-integrals)

## Key takeaways

- If sine or cosine has an odd power, save one factor for $du$ and turn the rest into the other function with $\sin^2 x + \cos^2 x = 1$.
- If both powers are even, lower them with the half-angle identities.
- For $\sec^n x$ with $n$ even, save $\sec^2 x$ for $du$ with $u = \tan x$. For odd powers of $\tan x$ times $\sec x$, save $\sec x\tan x$ with $u = \sec x$.

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:

$$\sin^2 x + \cos^2 x = 1, \qquad 1 + \tan^2 x = \sec^2 x$$

$$\sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2}$$

## 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

| Problem | Answer |
| --- | --- |
| 1 | $\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ |
| 2 | $\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$ |
| 3 | $\dfrac{x}{8} - \dfrac{\sin 4x}{32} + C$ |
| 4 | $\dfrac{\sin^3 x}{3} + C$ |
| 5 | $\sin x - \dfrac{\sin^3 x}{3} + C$ |
| 6 | $-\cos x + \dfrac{\cos^3 x}{3} + C$ |
| 7 | $-\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C$ |
| 8 | $\tan x - x + C$ |
| 9 | $\tan x + \dfrac{\tan^3 x}{3} + C$ |
| 10 | $\dfrac{\sec^3 x}{3} - \sec x + C$ |
| 11 | $-\dfrac{\cos 4x}{8} - \dfrac{\cos 2x}{4} + C$ |
| 12 | $\dfrac{\pi}{2}$ |

## Step-by-step solutions

**1.** Use the half-angle identity:

$$\int\frac{1 - \cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$$

See the [integral of sin² x](/integrals/sin-squared-x) for more.

**2.** Same idea with a plus sign:

$$\int\frac{1 + \cos 2x}{2}\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$$

**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 $2x$:

$$\frac{1}{4}\int\frac{1 - \cos 4x}{2}\,dx = \frac{x}{8} - \frac{\sin 4x}{32} + C$$

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

$$\int u^2\,du = \frac{\sin^3 x}{3} + C$$

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

$$\int(1 - u^2)\,du = \sin x - \frac{\sin^3 x}{3} + C$$

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

$$-\int(1 - u^2)\,du = -\cos x + \frac{\cos^3 x}{3} + C$$

**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 $u = \cos x$:

$$-\int(u^2 - u^4)\,du = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$$

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

$$\int(\sec^2 x - 1)\,dx = \tan x - x + C$$

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

$$\int(1 + u^2)\,du = \tan x + \frac{\tan^3 x}{3} + C$$

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

$$\int(u^2 - 1)\,du = \frac{\sec^3 x}{3} - \sec x + C$$

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

$$\frac{1}{2}\int(\sin 4x + \sin 2x)\,dx = -\frac{\cos 4x}{8} - \frac{\cos 2x}{4} + C$$

**12.** From problem 1:

$$\left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^{\pi} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

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](/learn/calculus-2/integration-techniques/trig-integrals). See also the [integral of tan² x](/integrals/tan-squared-x) and the [integral of sec³ x](/integrals/sec-cubed-x). When you see $\sqrt{a^2 - x^2}$ or $x^2 + a^2$, move on to [trig substitution problems](/practice-problems/trig-substitution).

## Frequently asked questions

### How do you integrate sin squared x?

Use $\sin^2 x = \frac{1 - \cos 2x}{2}$. Then $\int\sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$. Problem 1 shows the steps.

### What do you do when both sine and cosine have odd powers?

Either choice works. Save one factor from whichever has the smaller odd power to keep the algebra short.

### Why do trig integral answers look different in different books?

Trig identities let the same function be written many ways, and the answers may differ by a constant. Differentiate both forms or compare them at one point to check they match.