Question

Solve : $\int \left(sinx{)}^{-11/3}\right(cosx{)}^{-1/3}dx$

Answer 1

$\int {\left(sinx\right)}^{-11/3}{\left(cosx\right)}^{-1/3}dx$

Substitute,

$z=tanx$

$dz={sec}^{2}xdx$

$=\int \frac{1}{\frac{{\left(sinx\right)}^{11/3}}{{\left(cosx\right)}^{11/3}}{\left(cosx\right)}^{11/3}{\left(cosx\right)}^{1/3}}dx$

$=\int \frac{1}{{\left(tanx\right)}^{11/3}{\cos }^{4}x}dx$

$=\int \frac{{\sec }^{2}xdx}{{\left(tanx\right)}^{11/3}{\cos }^{2}x}$

$=\int \frac{\left({\sec }^{2}x\right)\left({\sec }^{2}x\right)dx}{{\left(tanx\right)}^{11/3}}$

$=\int \frac{(1+z^2)}{z^{11/3}}dz$

$=\int (z^{-11/3}+z^{-5/3})dz$

$=\frac{z^{-11/3+1}}{-11/3+1}+\frac{z^{-5/3+1}}{-5/3+1}+C$

$=\frac{z^{-8/3}}{-8/3}+\frac{z^{-2/3}}{-2/3}+C$

$=\frac{-3}{{8\left(tanx\right)}^{8/3}}-\frac{3}{2{\left(tanx\right)}^{2/3}}+C$