Evaluate:

Question:

Evaluate: $\int \frac{1}{1+\cos 3 x} d x$

Solution:

Let $I=\int \frac{1}{1+\cos 3 x} d x$

$=\int \frac{1}{1+\cos 3 x} d x$

Now Multiply with Conjugate,

$=\int \frac{1}{1+\cos 3 x} \times \frac{1-\cos 3 x}{1-\cos 3 x} d x$

$=\int \frac{1-\cos 3 x}{1-\cos ^{2} 3 x} d x$

$=\int \frac{1-\cos 3 x}{\sin ^{2} 3 x} d x$

$=\int \frac{1}{\sin ^{2} 3 \mathrm{x}} \mathrm{dx}-\int \frac{\cos 3 \mathrm{x}}{\sin ^{2} 3 \mathrm{x}} \mathrm{dx}$

$=\int\left(\operatorname{cosec}^{2} 3 x-\operatorname{cosec} 3 x \cdot \cot 3 x\right) d x$

$=-\frac{\cot 3 x}{3}+\frac{\operatorname{cosec} 3 x}{3}$

$=-\frac{1}{3} \cdot \frac{\cos 3 x}{\sin 3 x}+\frac{1}{3} \cdot \frac{1}{\sin 3 x}$

Hence, $l=\frac{1-\cos 3 x}{3 \sin 3 x}+C$

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