Trigonometric Integrals

Overview

Central question: Which power and parity patterns suggest identities? You will use identities to expose a factor that can become a differential, and you will distinguish the odd-power case from the all-even case. An identity is not inserted to make an expression look more complicated. It is chosen because it changes the integral into a form that has an antiderivative.

Look first at $\int\sin^3(x)\cos^2(x)\,dx$. The power of sine is odd. One sine factor can be saved, while the remaining even power $\sin^2(x)$ can be converted with $\sin^2(x)=1-\cos^2(x)$. Predict which substitution will then finish the problem.

Save one factor when a power is odd

Rewrite the integrand as

$$\sin^2(x)\sin(x)\cos^2(x) =\left(1-\cos^2(x)\right)\sin(x)\cos^2(x).$$

Now $u=\cos(x)$ is justified because $du=-\sin(x)\,dx$. The remaining integral is polynomial:

$$\int\sin^3(x)\cos^2(x)\,dx =-\int\left(1-u^2\right)u^2\,du =-\frac{u^3}{3}+\frac{u^5}{5}+C.$$

Returning to the original variable gives

$$-\frac{\cos^3(x)}{3}+\frac{\cos^5(x)}{5}+C.$$

The sign in $du=-\sin(x)\,dx$ is easy to lose. Write it immediately when selecting $u$; do not repair signs by intuition at the end. Differentiate the final answer to verify both the powers and the sign.

Use half-angle identities when both powers are even

For $\int\sin^2(x)\cos^2(x)\,dx$, neither sine nor cosine has an odd power to save. The previous strategy cannot create a lone differential factor. Instead, use the double-angle identity

$$\sin(x)\cos(x)=\frac12\sin(2x).$$

Squaring gives

$$\sin^2(x)\cos^2(x)=\frac14\sin^2(2x) =\frac18\left(1-\cos(4x)\right).$$

The integral is now direct:

$$\int\sin^2(x)\cos^2(x)\,dx =\frac{x}{8}-\frac{\sin(4x)}{32}+C.$$

Contrast: choose the factor that matches the derivative

For $\int\sin^4(x)\cos^3(x)\,dx$, cosine has an odd power. Save one $\cos(x)$, convert $\cos^2(x)$ to $1-\sin^2(x)$, and choose $u=\sin(x)$. Do not save sine simply because it appears first; $d(\sin x)=\cos x\,dx$ determines the productive pairing.

Explain the identity choice

The Pythagorean identity is useful here because it changes an even power into an expression involving the other function. In the odd-sine example, saving one sine leaves an even sine power. Replacing $\sin^2x$ by $1-\cos^2x$ produces a polynomial in the same function chosen for $u$. This is the desired structural outcome: after $u=\cos x$, the remaining expression contains only powers of $u$ and $du$.

If instead you rewrite $\cos^2x$ as $1-\sin^2x$ in that example, the saved sine factor no longer pairs naturally with the chosen $u=\cos x$. The algebra may remain true, but it does not expose the differential relationship. An identity is selected for compatibility with the factor being saved, not because one identity is universally preferable.

For all-even powers, half-angle identities are needed because neither function supplies a single differential factor. The formulas

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

lower the powers and replace the original product by sums of directly integrable functions. The argument $2x$ requires attention when integrating: $\int\cos(2x)\,dx=\frac12\sin(2x)+C$. The factor $\frac12$ comes from the derivative of the inside function, just as it did in substitution.

A definite-integral comparison

Consider $\int_0^{\pi/2}\sin^3x\cos^2x\,dx$. The same odd-power strategy applies, but the transformed bounds change when $u=\cos x$. At $x=0$, $u=1$; at $x=\frac\pi2$, $u=0$. The negative differential reverses the order, giving a positive integral. This is a useful combined retrieval: parity chooses the rewrite, substitution changes the variable, and the bounds must follow the new variable.

The final value should be positive because the original integrand is nonnegative on the stated interval. Checking sign and scale against the graph is especially useful when a minus sign enters through $du=-\sin x\,dx$.

Stop and check

A common incorrect start is to substitute $u=\sin(x)$ in $\int\sin^3(x)\cos^2(x)\,dx$ without first creating a lone $\cos(x)\,dx$. The misconception is choosing a familiar function without checking whether its differential is present. The correction is to inspect parity before choosing $u$ and to show the identity rewrite that creates the required differential.

Retrieval check

Set up, but do not fully simplify, $\int\cos^5(x)\sin^4(x)\,dx$. Because cosine has an odd power, save one $\cos(x)$, rewrite the remaining $\cos^4(x)$ as $\left(1-\sin^2(x)\right)^2$, and use $u=\sin(x)$. Explain why a half-angle identity is unnecessary in this case.

The next lesson uses a different trigonometric rewrite. Instead of powers of sine and cosine, radicals such as $\sqrt{a^2-x^2}$ are matched to a Pythagorean identity through a trigonometric substitution.