Test Index

UPSC NDA I and II 2020 Math Solved Paper

© examsnet.com

Question : 64 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:

Concept:

Use the identity

1+\tan^{2}x = \sec^{2}x

to simplify the integrand, then substitute

t = \tan x

.

Explanation:

Let

I = \int_{0}^{\pi/4} (\tan^{3}x + \tan x) \, dx

.

Factor

\tan x

:

I = \int_{0}^{\pi/4} \tan x (\tan^{2}x + 1) \, dx

.

Using

1+\tan^{2}x = \sec^{2}x

, we get

I = \int_{0}^{\pi/4} \tan x \sec^{2}x \, dx

.

Substitute

t = \tan x

. Then

dt = \sec^{2}x \, dx

.

When

x = 0

,

t = 0

; when

x = \pi/4

,

t = 1

.

Thus

I = \int_{0}^{1} t \, dt = \left[ \frac{t^{2}}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}

.

Answer:

\frac{1}{2}

(Option B).

© examsnet.com

Go to Question:

Prev Question

Next Question