Question

Evaluate: $\int\frac{1}{1+3sin^{2}x + 8cos^{2}x}dx$

• A. $\frac{1}{6}tan^{-1}\left(2tan x\right) +C$
• B. $tan^{-1}\left(2tan x\right) +C$
• C. $\frac{1}{6}tan^{-1}\left(\frac{2tan x}{3}\right) +C$
• D. None of these

Answer 1

Answer: (C)

$I = \int\frac{1}{1+3sin^{2}x + 8cos^{2}x}dx$

Dividing the numerator and denominator by$ cos^{2} x$, we get

$\Rightarrow I =\int\frac{sec ^{2}x}{sec^{2}x +3tan^{2}x +8}dx$

$\Rightarrow I =\int\frac{sec ^{2}x}{1+tan^{2}x +3tan^{2}x +8}dx = \int \frac{sec ^{2}x}{4tan^{2}x +9}dx$

putting $tan x= t \Rightarrow sec^{2}x dx = dt$, we get

$ I = \int\frac{dt}{4t^{2}+9} = \frac{1}{4} \int\frac{dt}{t^{2}+\left(\frac{3}{2}\right)^{2}} = \frac{1}{4}\times\frac{1}{\frac{3}{2}} tan ^{-1}\left(\frac{t}{\frac{3}{2}}\right) +C$

$\Rightarrow I= \frac{1}{6}tan^{-1}\left(\frac{2t}{3}\right)+C = \frac{1}{6}tan^{-1}\left(\frac{2tan x}{3}\right) +C$