Question

If $\int \frac{1}{25{\cos }^{2}x+9{\sin }^{2}x}dx=\frac{1}{A}{\tan }^{-1}\left(B\tan x\right)+C,$ then the value of $AB$ is
(where $A,B$ are fixed constants and $C$ is integration constant)

• A. $\frac{1}{25}$
• B. $25$
• C. $9$
• D. $\frac{1}{9}$

Answer 1

The correct option is C $9$

$I=\int \frac{1}{25{\cos }^{2}x+9{\sin }^{2}x}dx$
Dividing numerator and denominator by ${\cos }^{2}x$
$\Rightarrow I=\int \frac{{\sec }^{2}x}{25+9{\tan }^{2}x}dx$
Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt\Rightarrow I=\int \frac{dt}{(5{)}^2+(3t{)}^2}\Rightarrow I=\frac{1}{3}\cdot \frac{1}{5}{\tan }^{-1}\left(\frac{3t}{5}\right)+C[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]\Rightarrow I=\frac{1}{15}{\tan }^{-1}\left(\frac{3\tan x}{5}\right)+C\Rightarrow A=15,B=\frac{3}{5}\therefore AB=9$