Correct Answer is : log⁡(sec⁡x+tan⁡x)−tan⁡(x2)+c ( x+ x)- ≤ft(x/2)+clog(secx+tanx)−tan(2x)+c, where c is the constant of integration

( x+ x)+2 ≤ft(x/2)+c, where c is the constant of integration

( x+ x)-2 ≤ft(x/2)+c, where c is the constant of integration

( x+ x)+ ≤ft(x/2)+c, where c is the constant of integration

( x+ x)- ≤ft(x/2)+c, where c is the constant of integration

Correct Answer is : log⁡(sec⁡x+tan⁡x)−tan⁡(x2)+c ( x+ x)- ≤ft(x/2)+clog(secx+tanx)−tan(2x)+c, where c is the constant of integration

Test Index

MHTCET 5 May 2025 Shift 2 Paper

Question : 118 of 150

Marks: +1, -0

\int \frac{d x}{\cos x(1+\cos x)}=

$\quad \log (\sec x+\tan x)+2 \tan \left(\frac{x}{2}\right)+c$ , where c is the constant of integration
$\quad \log (\sec x+\tan x)-2 \tan \left(\frac{x}{2}\right)+c$ , where c is the constant of integration
$\log (\sec x+\tan x)+\tan \left(\frac{x}{2}\right)+c$ , where c is the constant of integration
$\log (\sec x+\tan x)-\tan \left(\frac{x}{2}\right)+c$ , where c is the constant of integration

Solution:

Concept:

Use partial fraction decomposition and standard integrals:

\int \sec x\,dx = \log|\sec x+\tan x|+c

and

\int \sec^2\frac{x}{2}\,dx = 2\tan\frac{x}{2}+c

Explanation:

Rewrite the integrand as

\frac{1}{\cos x(1+\cos x)} = \frac{1}{\cos x} - \frac{1}{1+\cos x}.

Therefore,

I = \int \sec x\,dx - \int \frac{1}{1+\cos x}\,dx.

Use the identity

1+\cos x = 2\cos^2\frac{x}{2}

to get

I = \int \sec x\,dx - \int \frac{1}{2}\sec^2\frac{x}{2}\,dx.

Evaluate each integral:

\int \sec x\,dx = \log|\sec x+\tan x|,

and

\int \frac{1}{2}\sec^2\frac{x}{2}\,dx = \tan\frac{x}{2}.

Thus,

I = \log|\sec x+\tan x| - \tan\frac{x}{2} + c.

Answer:

Option D:

\log(\sec x+\tan x)-\tan\left(\frac{x}{2}\right)+c

, where

c

is the constant of integration.

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