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TS EAMCET 4 May 2018 Shift 1 Solved Paper

Section: Mathematics

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Question : 73 of 160

Marks: +1, -0

\int\limits_{0}^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\sqrt{2018}}}

=

π
$\frac{3\pi}{4}$
$\frac{\pi}{4}$
${π}/4$

Solution:

Consider the integral.

I=\int\limits_{0}^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\sqrt{2018}}}

The above integral is solved as

I=\int\limits_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{\sqrt{2018}}}{(\cos x)^{\sqrt{2018}}+(\sin x)^{\sqrt{2018}}} dx

…… (I)

Also,

I=\int\limits_{0}^{\frac{\pi}{2}} \frac{(\sin x)^{\sqrt{2018}}}{(\sin x)^{\sqrt{2018}}+(\cos x)^{\sqrt{2018}}} dx

……. (II)

Add equation (I) and (II).

2I=\int\limits_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{\sqrt{2018}}+(\sin x)^{\sqrt{2018}}}{(\cos x)^{\sqrt{2018}}+(\sin x)^{\sqrt{2018}}} dx

2I=\int\limits_{0}^{\frac{\pi}{2}} dx

I=\frac{[x]_{0}^{\frac{\pi}{2}}}{2}

=\frac{\pi}{4}

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