Question

If $\int \frac{1}{(\sin x+4)(\sin x-1)}dx$
$=A\frac{1}{\left(\tan \frac{x}{2}-1\right)}+B{\tan }^{-1}[f(x)]+C_1$, 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,
$I=\int \frac{1}{(\sin x+4)(\sin x-1)}dx$
$=\frac{1}{5}\int \frac{(\sin x+4)-(\sin x-1)}{(\sin x+4)(\sin x-1)}dx$
$=\frac{1}{5}\int \frac{1}{\sin x-1}dx-\frac{1}{5}\int \frac{1}{\sin x+4}dx$
$=\frac{1}{5}\int \frac{{\sec }^2\frac{x}{2}}{2\tan \frac{x}{2}-1-{\tan }^2\frac{x}{2}}dx$
$-\frac{1}{5}\int \frac{{\sec }^2\frac{x}{2}}{2\tan \frac{x}{2}+4+4{\tan }^2\frac{x}{2}}dx$
Put, $\tan \frac{x}{2}=t$
$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$
$\therefore I=\frac{1}{5}\int \frac{2dt}{2t-1-t^2}-\frac{1}{5}\int \frac{2dt}{\left[2t+4\left(1+t^2\right)\right]}$
$\therefore I=-\frac{2}{5}\int \frac{dt}{t^2-2t+1}-\frac{1}{10}\int \frac{dt}{t^2+\frac{1}{2}t+1}$
$=\frac{-2}{5}\int \frac{1}{(t-1{)}^2}dt-\frac{1}{10}\int \frac{dt}{{\left(t+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{15}}{4}\right)}^2}$
$=\frac{2}{5}\cdot \frac{1}{t-1}-\frac{2}{5\sqrt{15}}{\tan }^{-1}\left(\frac{4t+1}{\sqrt{15}}\right)+C$
$=\frac{2}{5}\frac{1}{\tan \frac{x}{2}-1}-\frac{2}{5\sqrt{15}}{\tan }^{-1}\left(\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}\right)+C\dots \text{ (i) }$
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
$A=\frac{2}{5},B=\frac{-2}{5\sqrt{15}},f(x)=\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}$