# How do you find the indefinite integral of ∫ dx / sin^3 x cos^5 x ?

Apr 15, 2018

$I = \frac{1}{4} {\tan}^{4} \left(x\right) + \frac{3}{2} {\tan}^{2} \left(x\right) + 3 \ln \left(\tan \left(x\right)\right) - \frac{1}{2} {\cot}^{2} \left(x\right) + C$

#### Explanation:

We want to solve

$I = \int \frac{1}{{\sin}^{3} \left(x\right) {\cos}^{5} \left(x\right)} \mathrm{dx}$

Multiply the DEN and NUM by ${\sec}^{8} \left(x\right)$,
an integral consisting of tangens and secant have good opputunities substitution

$I = \int {\sec}^{8} \frac{x}{{\tan}^{3} \left(x\right)} \mathrm{dx}$

Make a substitution color(red)(u=tan(x)=>du=sec^2(x)dx

$I = \int {\sec}^{6} \frac{x}{{u}^{3}} \mathrm{du}$

Use the identity color(blue)(sec^2(x)=tan^2(x)+1

$I = \int {\left({u}^{2} + 1\right)}^{3} / \left({u}^{3}\right) \mathrm{du}$

$\textcolor{w h i t e}{I} = \int \frac{{u}^{6} + 3 {u}^{4} + 3 {u}^{2} + 1}{{u}^{3}} \mathrm{du}$

$\textcolor{w h i t e}{I} = \int {u}^{3} + 3 u + \frac{3}{u} + \frac{1}{u} ^ 3 \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{4} {u}^{4} + \frac{3}{2} {u}^{2} + 3 \ln \left(u\right) - \frac{1}{2 {u}^{2}} + C$

Substitute back $u = \tan \left(x\right)$

$I = \frac{1}{4} {\tan}^{4} \left(x\right) + \frac{3}{2} {\tan}^{2} \left(x\right) + 3 \ln \left(\tan \left(x\right)\right) - \frac{1}{2} {\cot}^{2} \left(x\right) + C$