Question

If $\int \frac{{\cos }^{3}xdx}{({\sin }^{4}x+3{\sin }^{2}x+1){\tan }^{-1}(\sin x+cosecx)}=-A$ $\log \left|{\tan }^{-1}(\sin x+cosecx)\right|+C$ then A is equal to.

Answer 1

$I=\int \frac{{\cos }^{3}x}{({\sin }^{4}x+3{\sin }^{2}x+1){\tan }^{-1}(\sin x+\csc x)}$

Substitute $t=\sin x$

$I=\int \frac{(1-t^2)dt}{(t^4+3t^2+1){\tan }^{-1}(t+\frac{1}{t})}=\int \frac{-(1-\frac{1}{t^2})dt}{(t^2+\frac{1}{t^2}+3){\tan }^{-1}(t+\frac{1}{t})}$

Again substitute $u=t+\frac{1}{t}$

$I=\int \frac{-du}{(u^2+1){\tan }^{-1}u}=-\log \left|{\tan }^{-1}u\right|+c$

$I=-\log \left|{\tan }^{-1}(\sin x+\csc x)\right|+c$

Comparing with, $\int \frac{{\cos }^{3}xdx}{({\sin }^{4}x+3{\sin }^{2}x+1){\tan }^{-1}(\sin x+cosecx)}=-A$ $\log \left|{\tan }^{-1}(\sin x+cosecx)\right|+C$

Therefore $A=1$