& = 4 ^3 -3 d & = 1 4 ^2 -3 d Divide Numerator and Denominator by ^2 & I= 1 ^2 4 ^2 -3 ^2 d & = ^2 4-3 ^2 d & = ^2 4-3 (1+ ^2 ) d & = ^2 1-3 ^2 d & put =t ^2 d =1 d t & I= 1 1-3 t^2 d t & =1 3 1 1 3 -t^2 d t & =1 3 1 (1 3 )^2-t^2 d t & =1 3 1 2 (1 3 ) ( 1 3 +t 1 3 -t )+c & =1 2 3 ( 1+3 t 1-3 t )+c & =1 2 3 ( 1+3 1-3 )+c & 3 d =1 2 3 ( 1+3 1-3 )+c

Evaluate : 3 d

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Question

Evaluate : $\int \frac{\cos \theta}{\cos 3 \theta} \cdot d \theta$

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Answer

$\begin{aligned}
& =\int \frac{\cos \theta}{4 \cos ^3 \theta-3 \cos \theta} \cdot d \theta \\
& =\int \frac{1}{4 \cos ^2 \theta-3} \cdot d \theta
\end{aligned}
$
Divide Numerator and Denominator by $\cos ^2 \theta$
$
\begin{aligned}
& I=\int \frac{\frac{1}{\cos ^2 \theta}}{\frac{4 \cos ^2 \theta-3}{\cos ^2 \theta}} \cdot d \theta \\
& =\int \frac{\sec ^2 \theta}{4-3 \sec ^2 \theta} \cdot d \theta \\
& =\int \frac{\sec ^2 \theta}{4-3\left(1+\tan ^2 \theta\right)} \cdot d \theta \\
& =\int \frac{\sec ^2 \theta}{1-3 \tan ^2 \theta} \cdot d \theta \\
& \text { put } \tan \theta=t \therefore \sec ^2 \theta \cdot d \theta=1 \cdot d t \\
& \mathrm{I}=\int \frac{1}{1-3 t^2} \cdot d t \\
& =\frac{1}{3} \cdot \int \frac{1}{\frac{1}{3}-t^2} \cdot d t \\
& =\frac{1}{3} \cdot \int \frac{1}{\left(\frac{1}{\sqrt{3}}\right)^2-t^2} \cdot d t \\
& =\frac{1}{3} \cdot \frac{1}{2\left(\frac{1}{\sqrt{3}}\right)} \cdot \log \left(\frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right)+c \\
& =\frac{1}{2 \sqrt{3}} \cdot \log \left(\frac{1+\sqrt{3} t}{1-\sqrt{3} t}\right)+c \\
& =\frac{1}{2 \sqrt{3}} \cdot \log \left(\frac{1+\sqrt{3} \tan \theta}{1-\sqrt{3} \tan \theta}\right)+c \\
& \therefore \int \frac{\cos \theta}{\cos 3 \theta} \cdot d \theta=\frac{1}{2 \sqrt{3}} \cdot \log \left(\frac{1+\sqrt{3} \tan \theta}{1-\sqrt{3} \tan \theta}\right)+c \\
\end{aligned}
$

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