Question

If $\int \frac{1}{(\sin x+4)(\sin x-1)} d x$
$=A \frac{1}{\left(\tan \frac{x}{2}-1\right)}+B \tan ^{-1}[f(x)]+C_1$, then

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

Answer 1

Answer: (D)

We have,
$ I=\int \frac{1}{(\sin x+4)(\sin x-1)} d x $
$=\frac{1}{5} \int \frac{(\sin x+4)-(\sin x-1)}{(\sin x+4)(\sin x-1)} d x $
$=\frac{1}{5} \int \frac{1}{\sin x-1} d x-\frac{1}{5} \int \frac{1}{\sin x+4} d x$
$=\frac{1}{5} \int \frac{\sec ^2 \frac{x}{2}}{2 \tan \frac{x}{2}-1-\tan ^2 \frac{x}{2}} d x$
$-\frac{1}{5} \int \frac{\sec ^2 \frac{x}{2}}{2 \tan \frac{x}{2}+4+4 \tan ^2 \frac{x}{2}} d x$
Put, $\tan \frac{x}{2}=t$
$\Rightarrow \sec ^2 \frac{x}{2} d x=2 d t$
$\therefore I=\frac{1}{5} \int \frac{2 d t}{2 t-1-t^2}-\frac{1}{5} \int \frac{2 d t}{\left[2 t+4\left(1+t^2\right)\right]}$
$\therefore I=-\frac{2}{5} \int \frac{d t}{t^2-2 t+1}-\frac{1}{10} \int \frac{d t}{t^2+\frac{1}{2} t+1}$
$=\frac{-2}{5} \int \frac{1}{(t-1)^2} d t-\frac{1}{10} \int \frac{d t}{\left(t+\frac{1}{4}\right)^2+\left(\frac{\sqrt{15}}{4}\right)^2}$
$=\frac{2}{5} \cdot \frac{1}{t-1}-\frac{2}{5 \sqrt{15}} \tan ^{-1}\left(\frac{4 t+1}{\sqrt{15}}\right)+C$
$=\frac{2}{5} \frac{1}{\tan \frac{x}{2}-1}-\frac{2}{5 \sqrt{15}} \tan ^{-1}\left(\frac{4 \tan \frac{x}{2}+1}{\sqrt{15}}\right)+C \ldots \text { (i) }$
But, given that
$I-A \frac{1}{\left(\tan \frac{x}{2}-1\right)}+B \tan ^{-1}[f(x)]+C$....(ii)
From Eqs. (i) and (ii), we get
$A=\frac{2}{5}, B=\frac{-2}{5 \sqrt{15}}, f(x)=\frac{4 \tan \frac{x}{2}+1}{\sqrt{15}}$