Mathematics

Evaluate:
$$\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { dx }{ \left( 3+2\cos { x }  \right)  }  }$$


SOLUTION

Consider the given integral.


$$I=\displaystyle \int_{0}^{\pi /2}{\dfrac{1}{3+2\cos x}dx}$$


$$ I=\displaystyle \int_{0}^{\pi /2}{\dfrac{1}{3+2\left( 1-2{{\sin }^{2}}\dfrac{x}{2} \right)}dx} $$


$$ I=\displaystyle \int_{0}^{\pi /2}{\dfrac{1}{3+2-4{{\sin }^{2}}\dfrac{x}{2}}dx} $$


$$ I=\displaystyle \int_{0}^{\pi /2}{\dfrac{{{\sec }^{2}}\dfrac{x}{2}}{5{{\sec }^{2}}\dfrac{x}{2}-4{{\tan }^{2}}\dfrac{x}{2}}dx} $$


$$ I=\displaystyle \int_{0}^{\pi /2}{\dfrac{{{\sec }^{2}}\dfrac{x}{2}}{5\left( {{\tan }^{2}}\dfrac{x}{2}+1 \right)-4{{\tan }^{2}}\dfrac{x}{2}}dx} $$


$$ I=\displaystyle \int_{0}^{\pi /2}{\dfrac{{{\sec }^{2}}\dfrac{x}{2}}{5+{{\tan }^{2}}\dfrac{x}{2}}dx} $$


 


Let $$t=\tan \dfrac{x}{2}$$


$$ \dfrac{dt}{dx}={{\sec }^{2}}\dfrac{x}{2}\left( \dfrac{1}{2} \right) $$


$$ 2dt={{\sec }^{2}}\dfrac{x}{2}dx $$


 


Therefore,


$$ I=2\displaystyle \int_{0}^{1}{\dfrac{1}{{{\left( \sqrt{5} \right)}^{2}}+{{t}^{2}}}dt} $$


$$ I=2\left[ \dfrac{1}{\sqrt{5}}{{\tan }^{-1}}\left( \dfrac{t}{\sqrt{5}} \right) \right]_{0}^{1} $$


$$ I=2\left[ \dfrac{1}{\sqrt{5}}{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right)-\dfrac{1}{\sqrt{5}}{{\tan }^{-1}}\left( 0 \right) \right] $$


$$ I=2\left[ \dfrac{1}{\sqrt{5}}{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right)-0 \right] $$


$$ I=\dfrac{2}{\sqrt{5}}{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right) $$


 


Hence, this is the answer.

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