The value of { ³ x}{≤ft( ^{4} x+3 ² x+1) ^{-1}( x+ x)} d x is

Correct Answer is : log⁡(tan⁡−1(sec⁡x+cos⁡x))+c ≤ft( ^{-1}( x+ x))+clog(tan−1(secx+cosx))+c

Correct Answer is : log⁡(tan⁡−1(sec⁡x+cos⁡x))+c ≤ft( ^{-1}( x+ x))+clog(tan−1(secx+cosx))+c

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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 129 of 150

Marks: +1, -0

The value of

\int \frac{\sin ^{3} x}{\left(\cos ^{4} x+3 \cos ^{2} x+1\right) \tan ^{-1}(\sec x+\cos x)} d x

Solution:

Concept:

Use substitution

u=\tan ^{-1}(\sec x+\cos x)

because its derivative appears in the numerator.

Explanation:

Let

u = \tan ^{-1}(\sec x+\cos x)

Differentiate

u

with respect to

x

\frac{du}{dx}=\frac{1}{1+(\sec x+\cos x)^2}\cdot \frac{d}{dx}(\sec x+\cos x)

Now,

\frac{d}{dx}(\sec x+\cos x)=\sec x\tan x-\sin x

Using

\sec x=\frac{1}{\cos x}

and

\tan x=\frac{\sin x}{\cos x}

\sec x\tan x-\sin x=\frac{\sin x}{\cos^2 x}-\sin x=\frac{\sin^3 x}{\cos^2 x}

Also,

1+(\sec x+\cos x)^2=1+\sec^2 x+2\sec x\cos x+\cos^2 x

Since

\sec x\cos x=1

1+(\sec x+\cos x)^2=\sec^2 x+\cos^2 x+3

=\frac{1}{\cos^2 x}+\cos^2 x+3=\frac{\cos^4 x+3\cos^2 x+1}{\cos^2 x}

Therefore,

\frac{du}{dx}=\frac{\frac{\sin^3 x}{\cos^2 x}}{\frac{\cos^4 x+3\cos^2 x+1}{\cos^2 x}}

=\frac{\sin^3 x}{\cos^4 x+3\cos^2 x+1}

So the given integral becomes:

\int \frac{\sin^3 x}{(\cos^4 x+3\cos^2 x+1)\tan^{-1}(\sec x+\cos x)}\,dx

=\int \frac{1}{u}\frac{du}{dx}\,dx

=\int \frac{1}{u}\,du=\log |u|+C

Thus,

=\log\left|\tan^{-1}(\sec x+\cos x)\right|+C

This matches Option A.

Answer:

Option A:

\log\left(\tan^{-1}(\sec x+\cos x)\right)+c

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