Q. 54 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int {\tan }^{3}x{\sec }^{2}x\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int {\tan }^3{\mathrm{xsec}}^2\mathrm{xdx}=\frac{1}{4}{\tan }^4\mathrm{x}+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals.

$\int {\tan }^{3}x{\sec }^{2}x\mathrm{d}x$

Step 2. Using the substitution method.

Let

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

Step 3. This substitution changes the integral into

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