Question
Evaluate : $\int \frac{1}{3+2 \sin ^2 x+5 \cos ^2 x} \cdot d x$
✓
Answer
Divide Numerator and Denominator by $\cos ^2 x$
$
\begin{aligned}
& I=\int \frac{\frac{1}{\cos ^2 x}}{\frac{3+2 \sin ^2 x+5 \cos ^2 x}{\cos ^2 x}} \cdot d x \\
& =\int \frac{\sec ^2 x}{3 \sec ^2 x+2 \tan ^2 x+5} \cdot d x \\
& =\int \frac{\sec ^2 x}{3\left(1+\tan ^2 x\right)+2 \tan ^2 x+5} \cdot d x \\
& =\int \frac{\sec ^2 x}{5 \tan ^2 x+8} \cdot d x \\
& =\frac{1}{5} \cdot \int \frac{\sec ^2 x}{\tan ^2 x+\frac{8}{5}} \cdot d x \\
& \text { put } \tan x=t \therefore \sec ^2 x \cdot d x=1 \cdot d t \\
& \mathrm{I}=\frac{1}{5} \cdot \int \frac{1}{t^2+\frac{8}{5}} \cdot d t \\
& =\frac{1}{5} \cdot \int \frac{1}{t^2+\left(\frac{\sqrt{8}}{\sqrt{5}}\right)^2} \cdot d t \\
& =\frac{1}{5} \cdot \frac{1}{\frac{\sqrt{8}}{\sqrt{5}}} \cdot \tan ^{-1}\left(\frac{t}{\frac{\sqrt{8}}{\sqrt{5}}}\right)+c \\
& =\frac{1}{\sqrt{5}} \cdot \frac{1}{2 \sqrt{2}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} t}{2 \sqrt{2}}\right)+c \\
& =\frac{1}{2 \sqrt{10}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} \tan x}{2 \sqrt{2}}\right)+c \\
& \therefore \int \frac{1}{3+2 \sin ^2 x+5 \cos ^2 x} \cdot d x= \\
& \frac{1}{2 \sqrt{10}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} \tan x}{2 \sqrt{2}}\right)+c \\
&
\end{aligned}
$
$
\begin{aligned}
& I=\int \frac{\frac{1}{\cos ^2 x}}{\frac{3+2 \sin ^2 x+5 \cos ^2 x}{\cos ^2 x}} \cdot d x \\
& =\int \frac{\sec ^2 x}{3 \sec ^2 x+2 \tan ^2 x+5} \cdot d x \\
& =\int \frac{\sec ^2 x}{3\left(1+\tan ^2 x\right)+2 \tan ^2 x+5} \cdot d x \\
& =\int \frac{\sec ^2 x}{5 \tan ^2 x+8} \cdot d x \\
& =\frac{1}{5} \cdot \int \frac{\sec ^2 x}{\tan ^2 x+\frac{8}{5}} \cdot d x \\
& \text { put } \tan x=t \therefore \sec ^2 x \cdot d x=1 \cdot d t \\
& \mathrm{I}=\frac{1}{5} \cdot \int \frac{1}{t^2+\frac{8}{5}} \cdot d t \\
& =\frac{1}{5} \cdot \int \frac{1}{t^2+\left(\frac{\sqrt{8}}{\sqrt{5}}\right)^2} \cdot d t \\
& =\frac{1}{5} \cdot \frac{1}{\frac{\sqrt{8}}{\sqrt{5}}} \cdot \tan ^{-1}\left(\frac{t}{\frac{\sqrt{8}}{\sqrt{5}}}\right)+c \\
& =\frac{1}{\sqrt{5}} \cdot \frac{1}{2 \sqrt{2}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} t}{2 \sqrt{2}}\right)+c \\
& =\frac{1}{2 \sqrt{10}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} \tan x}{2 \sqrt{2}}\right)+c \\
& \therefore \int \frac{1}{3+2 \sin ^2 x+5 \cos ^2 x} \cdot d x= \\
& \frac{1}{2 \sqrt{10}} \cdot \tan ^{-1}\left(\frac{\sqrt{5} \tan x}{2 \sqrt{2}}\right)+c \\
&
\end{aligned}
$
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