The value of { ³ x}{ ² x+ x} d x is

Correct Answer is : log⁡(sin⁡x)−sin⁡x+c ( x)- x+clog(sinx)−sinx+c, where c is a constant of integration.

( x)- x+c, where c is a constant of integration.

( x)+ x+c, where c is a constant of integration.

( x)- x+c, where c is a constant of integration.

Correct Answer is : log⁡(sin⁡x)−sin⁡x+c ( x)- x+clog(sinx)−sinx+c, where c is a constant of integration.

Test Index

MHTCET 11 May 2024 Shift 1 Paper

Question : 145 of 150

Marks: +1, -0

The value of

\int \frac{\cos ^{3} x}{\sin ^{2} x+\sin x} d x

Solution:

Concept:

Simplify the integrand using trigonometric identities, then substitute

t = \sin x

to integrate easily.

Explanation:

Write

\cos^3 x = \cos x \cdot \cos^2 x = \cos x(1 - \sin^2 x)

So,

I = \int \frac{\cos x(1 - \sin^2 x)}{\sin^2 x + \sin x}\,dx

Factorize:

\sin^2 x + \sin x = \sin x(1 + \sin x)

and

1 - \sin^2 x = (1 - \sin x)(1 + \sin x)

Cancel

(1 + \sin x)

from numerator and denominator.

Thus,

I = \int \frac{\cos x(1 - \sin x)}{\sin x}\,dx

Let

t = \sin x

, then

dt = \cos x\,dx

I = \int \frac{1 - t}{t}\,dt = \int \left(\frac{1}{t} - 1\right) dt

Integrate:

I = \log|t| - t + c

Substitute back

t = \sin x

I = \log|\sin x| - \sin x + c

Answer:

\log(\sin x) - \sin x + c

, where

c

is a constant of integration.

Correct option: A.

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