Question

Solve: $\int ta{n}^{3}xdx.$

Answer 1

$\int {\tan }^{3}xdx=\int \tan x.{\tan }^{2}xdx=\int \tan x({\sec }^{2}x-1)dx=\int {\sec }^{2}x\tan xdx-\int \tan xdx=I_1-I_2$

Now, $I_1=\int {\sec }^{2}x\tan xdx$

Let $\tan x=t{\sec }^{2}xdx=dt$

$I_1=\int {\sec }^{2}x\tan xdx=\int tdt=\frac{t^2}{2}+c_1=\frac{{\tan }^{2}x}{2}+c_1{I}_2=\int \tan xdx=-\log |\cos x|+c_2$

Hence, $I_1-I_2=\frac{{\tan }^{2}x}{2}+c_1+\log |\cos x|+c_2=\frac{{\tan }^{2}x}{2}+\log |\cos x|+c$