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Question
` ∫ 1/ {1+ cos 3x} ` dx
Sum
Solution
` ∫ 1/ {1+ cos 3x} ` dx
\[ = \int\frac{\left( 1 - \cos 3x \right)}{\left( 1 + \text{cos 3x} \right) \left( 1 - \cos 3x \right)}dx\]
\[ = \int\left( \frac{1 - \cos 3x}{1 - \cos^2 3x} \right) dx\]
\[ = \int\left( \frac{1 - \cos 3x}{\sin^2 3x} \right) dx\]
\[ = \int \text{cosec}^\text{2}\text{ 3x dx} - ∫cosec\ 3x \cot 3xdx\]
` = - {cot 3x} / 3 + {"cosec " 3x} / 3 + c `
` = 1/3 [ "cosec" 3x - cot 3x ] + c `
\[ = \frac{1}{3}\left[ \frac{1}{\sin 3x} - \frac{\cos 3x}{\sin 3x} \right] + C\]
\[ = \frac{1}{3} \left[ \frac{1 - \cos 3x}{\sin 3x} \right] + C\]
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