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
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