Question

If $\int \frac{({\cos }^{3}x+{\cos }^{5}x)}{({\sin }^{2}x+{\sin }^{4}x)}dx=A\sin x+B\text{cosec}x+C{\tan }^{-1}\left(\sin x\right)+K,$ where $A,B,C$ are fixed constants and $K$ is constant of integration, then the value of $A+B-C$ is equal to

• A. $7$
• B. $3$
• C. $5$
• D. $4$

Answer 1

The correct option is C $5$

Let
$I=\int \frac{({\cos }^{3}x+{\cos }^{5}x)}{({\sin }^{2}x+{\sin }^{4}x)}dx$
$\Rightarrow I=\int \frac{({\cos }^{2}x+{\cos }^{4}x)}{({\sin }^{2}x+{\sin }^{4}x)}\cos xdx$
Put $\sin x=t$
$\Rightarrow \cos xdx=dt$
$\therefore I=\int \frac{(1-t^2)+(1+t^4-2t^2)}{t^2+t^4}dt$
$\Rightarrow I=\int \frac{t^4-3t^2+2}{t^2(t^2+1)}dt$
$\Rightarrow I=\int [1+\frac{(2-4t^2)}{t^2(t^2+1)}]dt$
$\Rightarrow I=\int 1\cdot dt+2\int \frac{(t^2+1)-t^2}{t^2(t^2+1)}dt-4\int \frac{1}{t^2+1}dt$
$\Rightarrow I=\int 1\cdot dt+2\int \frac{1}{t^2}dt-6\int \frac{1}{t^2+1}dt$
$\Rightarrow I=t-\frac{2}{t}-6{\tan }^{-1}\left(t\right)+K$
$\Rightarrow I=\sin x-\frac{2}{\sin x}-6{\tan }^{-1}\left(\sin x\right)+K$
$\Rightarrow A=1,B=-2$ and $C=-6$
$\therefore A+B-C=5$