Question

Trigonometric integrals: The integrals that follow can be solved by using algebra to write the integrands in the form $f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)$so that$u-$substitution will apply.

Solve $\int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x$by using the Pythagorean identity role="math" localid="1649174356968" ${\tan }^{2}x+1={\sec }^{2}x$to rewrite the integrand as $({\tan }^{2}x+1){\tan }^{3}x{\sec }^{2}x$ and then applying substitution with $u=\tan x$.

Short answer

The value of integral is $\int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\frac{{\sec }^6\mathrm{x}}{6}+\frac{{\sec }^{4}x}{4}+C$.

Step-by-step solution

Step 1. Given Information 

Solve $\int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x$by using the Pythagorean identity ${\tan }^{2}x+1={\sec }^{2}x$to rewrite the integrand as $({\tan }^{2}x+1){\tan }^{3}x{\sec }^{2}x$ and then applying substitution with $u=\tan x$.

Step 2. The given integral is ∫sec4xtan3xdx

We can write as

$\int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\int ({\tan }^{2}x+1){\tan }^{3}x{\sec }^{2}xdx$

Let

$$\begin{gathered} u=\tan x \\ \frac{du}{dx}={\sec }^{2}x \\ du={\sec }^{2}xdx \end{gathered}$$

Step 3. Now the integral is 

$$\begin{gathered} \int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\int (u^2+1)u^{3}du \\ \int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\int (u^2·u^3+1·u^3)du \\ \int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\int (u^5+u^3)du \\ \int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\int u^{5}du+\int u^{3}du \\ \int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\left[\frac{u^{5+1}}{5+1}\right]+\left[\frac{u^{3+1}}{3+1}\right]+C \\ \int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\frac{u^6}{6}+\frac{u^4}{4}+C \\ \int {\sec }^{4}x{\tan }^{3}x\mathrm{d}x=\frac{{\sec }^6\mathrm{x}}{6}+\frac{{\sec }^{4}x}{4}+C \end{gathered}$$