Question

If $\int \frac{1}{(\sin x+4)(\sin x-1)}dx=A\frac{1}{\tan \frac{x}{2}-1}+B{\tan }^{-1}\left(f\right(x\left)\right)+c_1$
then

• A. $A=\frac{1}{5},B=\frac{-2}{5\sqrt{15}},f\left(x\right)=\frac{4\tan x+3}{\sqrt{15}}$
• B. $A=-\frac{1}{5},B=\frac{1}{\sqrt{15}},f\left(x\right)=\frac{4\tan \left(\frac{x}{2}\right)+1}{\sqrt{15}}$
• C. $A=\frac{2}{5},B=\frac{-2}{5},f\left(x\right)=\frac{4\tan x+1}{5}$
• D. $A=\frac{2}{5},B=\frac{-2}{5\sqrt{15}},f\left(x\right)=\frac{4\tan \left(\frac{x}{2}\right)+1}{\sqrt{15}}$

Answer 1

Answer: (D)

$A=\frac{2}{5},B=\frac{-2}{5\sqrt{15}},f\left(x\right)=\frac{4\tan \left(\frac{x}{2}\right)+1}{\sqrt{15}}$

Let $I=\int \frac{1}{(\sin x+4)(\sin x-1)}dx$
$=\frac{1}{5}\int \frac{(\sin x+4)-(\sin x-1)}{(\sin x+4)(\sin x-1)}dx$
$=\frac{1}{5}\int \frac{2dt}{2t-1-t^2}-\frac{1}{5}\int \frac{2dt}{2t-4(1+t^2)}[\tan \frac{x}{2}=t]$
$=-\frac{2}{5}\int \frac{dt}{2t-1-t^2}-\frac{1}{10}\int \frac{dt}{t^2+\frac{1}{2}t+1}$
$=\frac{2}{5}\frac{1}{(t-1)}-\frac{2}{5\sqrt{15}}{\tan }^{-1}\left(\frac{4t+1}{\sqrt{15}}\right)+C$
$=\frac{2}{5}\frac{1}{\tan \frac{x}{2}-1}-\frac{2}{5\sqrt{15}}{\tan }^{-1}\left(\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}\right)+C$
But $\int \frac{1}{(\sin x+4)(\sin x-1)}dx=A\frac{1}{\tan \frac{x}{2}-1}+B{\tan }^{-1}\left(f\right(x\left)\right)+C_1$
Therefore, $A=\frac{2}{5},B=-\frac{2}{5\sqrt{15}},f\left(x\right)=\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}$