Question

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

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

Answer 1

Answer: (D)

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

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

Breaking the LHS in partial fraction form we have ,

$\int \frac{dx}{(\sin x+4)(\sin x-1)}=\frac{1}{5}\int (\frac{dx}{(\sin x-1)}-\frac{dx}{(\sin x+4)})$

Now by breaking $\sin x$ in submultiple angle we have LHS as ,

$\Rightarrow$ $\frac{1}{5}\int (\frac{dx}{(2\sin \frac{x}{2}\cos \frac{x}{2}-1)}-\frac{dx}{(2\sin \frac{x}{2}\cos \frac{x}{2}+4)})$ ......$\left(1\right)$

$\Rightarrow$ Putting $\tan \frac{x}{2}=t$ we have,

$\Rightarrow$ $\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$............$\left(2\right)$

Putting $\left(2\right)$ in $\left(1\right)$ we get our LHS as,

$\Rightarrow$ $\frac{1}{5}\int (\frac{dt}{(2\sin \frac{x}{2}\cos \frac{x}{2}-1)}-\frac{dt}{(2\sin \frac{x}{2}\cos \frac{x}{2}+4)})\frac{2}{{\sec }^2\frac{x}{2}}$

$\Rightarrow$ $\frac{1}{5}\int (\frac{(-2)dt}{{(t-1)}^2}-\frac{(-2)dt}{(2t+4(1+t^2))})$

$\Rightarrow$ $\frac{1}{5}\int (\frac{(-2)dt}{{(t-1)}^2}-\frac{(-2)dt}{{(2t+\frac{1}{2})}^2+\frac{{\left(\sqrt{15}\right)}^2}{2^2}})$

$\Rightarrow$ $\frac{\frac{2}{5}}{\tan \frac{x}{2}-1}+\left(\frac{-2}{5\sqrt{15}}\right){\tan }^{-1}\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}+c$

Comparing the above with the very first equation we get

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