NIMCET 2023 Question Paper with solutions

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Question : 22 of 120

Marks: +1, -0

If

∫ x sin \;x \ sec ^3 x \dx=\;{1}/{2} [f(x) \ sec ^2 x+g(x) (\;{\tan x}/{x}) ] +c

, where

f

and

g

are real valued functions of

x

and

c

is an arbitrary constant, then which of the following statement is true?

$f(x)=g(x)$
$f(x) ⋅ g(x)=1$
$f(x)+g(x)=0$
$f(x)-g(x)=c$

Solution:

∫ x sin \;x \ sec ^3 x \dx

=∫ x \tan x \ sec ^2 x \dx

Using integration by part

\;= [x ∫ \ sec ^2 x \tan x \dx] -∫ (\;{d x}/{d x} ∫ \ sec ^2 x \tan x \dx) \dx

\;= [x \;{\tan ^2 x}/{2}] -∫ \;{\tan ^2 x}/{2} \dx

\;=\;{1}/{2} [x \tan ^2 x-∫ (\ sec ^2 x-1) \dx]

\;=\;{1}/{2} [x \tan ^2 x-\tan x+x] +c

\;=\;{1}/{2} [x (1+\tan ^2 x) -\tan x] +c

\;=\;{1}/{2} [x \ sec ^2 x-\tan x] +c

\;=\;{1}/{2} [f(x) \ sec ^2 x+g(x) (\;{\tan x}/{x}) ] +c

Hence,

f(x)=x: g(x)=-x

Hence,

f(x)+g(x)=0

.

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