# Evaluate the integral: {eq}\displaystyle \int \frac{1}{2 + 2\ cos\ \theta + sin\ \theta}\ d \theta {/eq}

## Question:

Evaluate the integral:

{eq}\displaystyle \int \frac{1}{2 + 2\ cos\ \theta + sin\ \theta}\ d \theta {/eq}

## Indefinite Integral:

If the integral in the form of {eq}\int {\dfrac{1}{{a\sin x + b\cos x}}dx} {/eq}, then the integral is evaluated by substituting, {eq}\cos x = \dfrac{{1 - {{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}} {/eq}, and {eq}\sin x = \dfrac{{2\tan x/2}}{{1 + {{\tan }^2}x/2}} {/eq}. To evaluate the integral further use the substitution method and substitute {eq}\tan \dfrac{x}{2} = u {/eq}, so that its derivative is {eq}\dfrac{1}{2}{\sec ^2}\dfrac{x}{2}dx = du {/eq}. The derivative of logarithm function {eq}\log x {/eq} is {eq}\dfrac{1}{x} {/eq}.

## Answer and Explanation: 1

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Given:

• The integral is, {eq}\int {\dfrac{1}{{2 + 2\cos \theta + \sin \theta }}d\theta } {/eq}.

As it is known that, from the identity,

{eq}\co...

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