Question

If $\int \frac{dx}{4-3{\cos }^{2}x+5{\sin }^{2}x}=Af\left(3\tan x\right)+C,$ for a combination of function $f$ and a fixed constant $A.$ Then which of the following is/are true
(where $C$ is integration constant)

• A. $f\left(x\right)=x^2+1$
• B. $A=\frac{1}{9}$
• C. $A=\frac{1}{3}$
• D. $f\left(x\right)={\tan }^{-1}x$

Answer 1

Answer: (C), (D)

$f\left(x\right)={\tan }^{-1}x$

$I=\int \frac{dx}{4-3{\cos }^{2}x+5{\sin }^{2}x}$
Dividing numerator by ${\cos }^{2}x$, the integral becomes
$\Rightarrow I=\int \frac{{\sec }^{2}xdx}{4(1+{\tan }^{2}x)-3+5{\tan }^{2}x}$
Putting
$\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow I=\int \frac{dt}{9t^2+1}\Rightarrow I=\frac{1}{9}\int \frac{dt}{t^2+{\left(\frac{1}{3}\right)}^2}\Rightarrow I=\frac{1}{3}{\tan }^{-1}\left(3t\right)+C[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]\therefore I=\frac{1}{3}{\tan }^{-1}\left(3\tan x\right)+C$
Thus,
$A=\frac{1}{3},f\left(x\right)={\tan }^{-1}x$