Question

use any method to evaluate the integral.
int\frac{\tan^{2}5x}{csc 5x}dx
int\frac{\tan^{2}5x}{csc 5x}dx=square

Explanation:

Step1: Rewrite trig - functions

Recall that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\csc\theta = \frac{1}{\sin\theta}$. So, $\frac{\tan^{2}5x}{\csc5x}=\frac{\frac{\sin^{2}5x}{\cos^{2}5x}}{\frac{1}{\sin5x}}=\frac{\sin^{3}5x}{\cos^{2}5x}$. Let $u = \cos5x$, then $du=- 5\sin5x dx$ and $\sin^{2}5x = 1 - \cos^{2}5x=1 - u^{2}$.

Step2: Substitute

The integral $\int\frac{\sin^{3}5x}{\cos^{2}5x}dx=\int\frac{\sin^{2}5x\cdot\sin5x}{\cos^{2}5x}dx$. Substituting $u = \cos5x$ and $\sin^{2}5x=1 - u^{2}$, we get $-\frac{1}{5}\int\frac{1 - u^{2}}{u^{2}}du=-\frac{1}{5}\int(\frac{1}{u^{2}}-1)du$.

Step3: Integrate term - by - term

$-\frac{1}{5}\int(\frac{1}{u^{2}}-1)du=-\frac{1}{5}\int(u^{-2}-1)du$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $-\frac{1}{5}(\frac{u^{-2 + 1}}{-2 + 1}-u)+C=-\frac{1}{5}(-\frac{1}{u}-u)+C$.

Step4: Back - substitute

Substituting back $u = \cos5x$, we get $\frac{1}{5\cos5x}+\frac{\cos5x}{5}+C=\frac{1}{5}\sec5x+\frac{1}{5}\cos5x+C$.

Answer:

$\frac{1}{5}\sec5x+\frac{1}{5}\cos5x + C$