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NDA Math Definite Integrals

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Question : 14 of 62

Marks: +1, -0

What is the value of

\int\limits_{0}^{\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) \cdot dx

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

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

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

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

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

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

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

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

Let

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

\therefore \int \tan x \cdot \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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