Question

If $I_1=\int \frac{\tan x}{a+b{\tan }^{2}x}dx=c_1\ln |f\left(x\right)|+c_2$
$I_2=\int \frac{\sin 2x}{a{\cos }^{2}x+b{\sin }^{2}x}dx,$ then which of the following (s) is/are correct

• A. $f\left(x\right)$ is a periodic function with period $\pi$
• B. $f\left(x\right)$ is a periodic function with period $\frac{\pi }{2}$
• C. $2I_1=I_2$
• D. $I_1=2I_2$

Answer 1

Answer: (A), (C)

$2I_1=I_2$

$I_1=\int \frac{\tan x}{a+b{\tan }^{2}x}dx=\int \frac{\frac{\sin x}{\cos x}}{a+b{\left(\frac{\sin x}{\cos x}\right)}^2}dx=\int \frac{\sin x.\cos x}{a{\cos }^{2}x+b{\sin }^{2}x}dx=\frac{1}{2}\int \frac{\sin 2xdx}{a{\cos }^{2}x+b{\sin }^{2}x}\Rightarrow I_1=\frac{1}{2}{I}_2\Rightarrow 2I_1=I_2$
Now, substituting $a{\cos }^{2}x+b{\sin }^{2}x=y$
$\Rightarrow 2(b-a)\sin x\cos xdx=dy\Rightarrow (b-a)\sin 2xdx=dy{I}_1=\frac{1}{2}\int \frac{1}{y}\cdot \frac{dy}{(b-a)}=\frac{1}{2(b-a)}\ln |y|+C=\frac{1}{2(b-a)}\ln |a{\cos }^{2}x+b{\sin }^{2}x|+C\Rightarrow f\left(x\right)=a{\cos }^{2}x+b{\sin }^{2}x$
$\Rightarrow f(x+\pi )=a{\cos }^2(x+\pi )+b{\sin }^2(x+\pi )=f\left(x\right)$
$\Rightarrow$ Periodic with period $\pi$