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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Subjective Medium Published on 17th 09, 2020
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