Mathematics

# Solve $\int\limits_0^{\pi /2} {\dfrac{{dx}}{{1 + {{\left( {\tan x} \right)}^{\sqrt 2 ,}}}}}$

##### SOLUTION

We have,

$I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{dx}{1+{{\left( \tan x \right)}^{\sqrt{2}}}}}$

$I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \cos x \right)}^{\sqrt{2}}}dx}{{{\left( \cos x \right)}^{\sqrt{2}}}+{{\left( \sin x \right)}^{\sqrt{2}}}}}$       …….. (1)

We know that,

$\int_{a}^{b}{f\left( x \right)}dx=\int_{a}^{b}{f\left( a+b-x \right)}dx$

Therefore,

$I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \cos \left( \dfrac{\pi }{2}-x \right) \right)}^{\sqrt{2}}}dx}{{{\left( \cos \left( \dfrac{\pi }{2}-x \right) \right)}^{\sqrt{2}}}+{{\left( \sin \left( \dfrac{\pi }{2}-x \right) \right)}^{\sqrt{2}}}}}$

$I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin x \right)}^{\sqrt{2}}}dx}{{{\left( \sin x \right)}^{\sqrt{2}}}+{{\left( \cos x \right)}^{\sqrt{2}}}}}$             ………. (2)

On adding equation (1) and (2), we get

$2I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\left( \sin x \right)}^{\sqrt{2}}}+{{\left( \cos x \right)}^{\sqrt{2}}}}{{{\left( \sin x \right)}^{\sqrt{2}}}+{{\left( \cos x \right)}^{\sqrt{2}}}}}dx$

$2I=\int_{0}^{\dfrac{\pi }{2}}{1}dx$

$2I=\left( x \right)_{0}^{\dfrac{\pi }{2}}$

$2I=\dfrac{\pi }{2}$

$I=\dfrac{\pi }{4}$

Hence, the value is $\dfrac{\pi }{4}$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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