Examples

Power product - odd power

Compute the integral:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx$$

Solution

(1) Swap over the even bunch:

Max even bunch leaving power-one is ${\sin }^{4}x$.

$${\sin }^{5}x\gg \gg \sin x({\sin }^{2}x{)}^2\gg \gg \sin x(1-{\cos }^{2}x{)}^2$$

Apply to ${\sin }^{5}x$ in the integrand:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx\gg \gg \int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx$$

---

(2) Perform $u$-substitution on the power-one integrand:

Set $u=\cos x$. Hence $du=\sin xdx$. Recognize this in the integrand and convert:

$$\begin{array}{cc}\int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx& \gg \gg \int {\cos }^{2}x\cdot (1{\cos }^{2}x{)}^2(\sin xdx)\\ & \\ & \gg \gg \int u^2\cdot (1-u^2{)}^{2}du\end{array}$$

---

(3) Integrate using power rule:

$$\int u^2\cdot {(1-u^2)}^{2}du=\frac{1}{3}{u}^3-\frac{2}{5}{u}^5+\frac{1}{7}{u}^7+C$$

Insert definition $u=\cos x$:

$$\begin{array}{c}\gg \gg \int {\cos }^{2}x\cdot {\sin }^{5}xdx\gg \gg \int u^2\cdot {(1-u^2)}^{2}du\\ \\ \gg \gg \frac{1}{3}{\cos }^{3}x-\frac{2}{5}{\cos }^{5}x+\frac{1}{7}{\cos }^{7}x+C\end{array}$$

Power product - tan and sec

Compute the integral:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx$$

Solution

(1) Try $du={\sec }^{2}xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{5}x\cdot \sec x({\sec }^{2}xdx)$$

We then must swap over remaining $\sec x$ into the $\tan x$ type.

Cannot do this because $\sec x$ has odd power. Need even to swap.

---

(2) Try again: $du=\sec x\tan xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{4}x\cdot {\sec }^{2}x\left(\sec x\tan xdx\right)$$

Swap remaining $\tan x$ into $\sec x$ type:

$$\begin{array}{c}\gg \gg \int ({\tan }^{2}x{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\\ \\ \gg \gg \int ({\sec }^{2}x-1{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\end{array}$$

Substitute $u=\sec x$ and $du=\sec x\tan xdx$:

$$\gg \gg \int {(u^2-1)}^2\cdot u^{2}du$$

---

(3) Integrate in $u$ and convert back to $x$:

$$\begin{array}{c}\gg \gg \int u^6-2u^4+u^{2}du\\ \\ \gg \gg \frac{u^7}{7}-2\frac{u^5}{5}+\frac{u^3}{3}+C\\ \\ \gg \gg \frac{{\sec }^{7}x}{7}-2\frac{{\sec }^{5}x}{5}+\frac{{\sec }^{3}x}{3}+C\end{array}$$

Trig power product - differing frequencies

Compute the integral:

$$\int \sin 4x\cdot \cos 5xdx$$

Solution

(1) Convert product to sum using trig identity:

Use $\sin A\cos B=\frac{1}{2}(\sin (A-B)+\sin (A+B))$ with $A=4x$ and $B=5x$:

$$\sin 4x\cdot \cos 5x\gg \gg \frac{1}{2}(\sin (-x)+\sin (9x))$$

---

(2) Integrate:

Break up the sum:

$$\int \sin 4x\cdot \cos 5xdx\gg \gg \frac{1}{2}\int \sin (-x)dx+\frac{1}{2}\int \sin (9x)dx$$

Observe chain rule backwards:

$$\frac{1}{2}\int \sin (-x)dx+\frac{1}{2}\int \sin (9x)\gg \gg \frac{1}{2}\cos (-x)-\frac{1}{18}\cos (9x)+C$$