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UPSC NDA I and II 2020 Math Solved Paper

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Question : 94 of 120

Marks: +1, -0

What is the value of

\int\limits_{0}^{\frac{\pi}{4}} (\tan^{3} x+\tan x) \, dx ?

$\frac{1}{4}$
$\frac{1}{2}$
$1$
$2$

Solution: 👈: Video Solution

I=\int\limits_{0}^{\frac{\pi}{4}} (\tan^{3} x+\tan x) .dx

=\int\limits_{0}^{\frac{\pi}{4}} \tan^{3} x .dx+\int\limits_{0}^{\frac{\pi}{4}} \tan x .dx

=\int\limits_{0}^{\frac{\pi}{4}} \tan x \tan^{2} x .dx+\int\limits_{0}^{\frac{\pi}{4}} \tan x .dx

=\int\limits_{0}^{\frac{\pi}{4}} \tan x .(\sec^{2}x-1) .dx

+\int\limits_{0}^{\frac{\pi}{4}} \tan x .dx

=\int\limits_{0}^{\frac{\pi}{4}} \tan x . \sec^{2}x.dx

-\int\limits_{0}^{\frac{\pi}{4}} \tan x .dx

+\int\limits_{0}^{\frac{\pi}{4}} \tan x .dx

=\int\limits_{0}^{\frac{\pi}{4}} \tan x . \sec^{2}x.dx

Let

\tan x=z,dx=\sec^{2}x.dx

\therefore \int \tan x . \sec^{2} x . dx=\int z . dz=\frac{z^{2}}{2}+C

where

C=

constant

\therefore I=\left[\frac{\tan^{2} x}{2}\right]_{0}^{\frac{\pi}{4}}

=\frac{1}{2}-0=\frac{1}{2}

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