Answer to Question #153390 in Calculus for Malik

Question

$\int$ Cosec⁶(2x) dx

Answer 1

Let

$I=\int cosec^6(2x)dx$

Let $y=2x\Rightarrow dy=2dx$

Now $I=\int cosec^6(2x)dx=\frac{1}{2}\int cosec^6(y)dy$

$\therefore 2I\\= \int cosec^6ydy\\ =\int cosec^4y\times cosec^2ydy\\ =\int cosec^4y\times(1+cot^2y)dy\\ =\int (cosec^4y)dy+\int cosec^4y\times cot^2ydy\\ =\int (cosec^2y\times(1+ cot^2y))dy+\int cosec^2y\times cot^2y\times (1+cot^2y)dy\\ =\int cosec^2ydy+\int cosec^2ycot^2ydy+\int cosec^2ycot^2ydy+\int cot^4ycosec^2ydy\\$

Let

$coty=z\\\Rightarrow -cosec^2ydy=dz\\\Rightarrow cosec^2ydy=-dz$

Now replacing $coty \;by\; z\;and \;cosec^2ydy\;by \;(-dz),$

$2I\\ =-\int dz-2\int z^2dz-\int z^4dz\\ =-z- \frac{2z^3}{3}-\frac{z^5}{5}+c'$

[ c' is the integrating constant ]

$=-cot(2x)-\frac{2}{3}cot^3(2x)-\frac{1}{5}cot^5(2x)+c'$

[ putting $z=coty$ and, $y=2x$ ]

$\Rightarrow I=-\frac{1}{2}cot(2x)-\frac{2}{6}cot^3(2x)-\frac{1}{10}cot^5(2x)+c$

[ $c=c'/2]$

So,

$$\boxed{I=\int cosec^6(2x)=-\frac{1}{2}cot(2x)-\frac{2}{6}cot^3(2x)-\frac{1}{10}cot^5(2x)+c}$$

where $c$ is a Integrating Constant.