If x [-1,1], then the value of e^{ ^{-1} x}≤ft({x+{1-x²}}{{1-x²}}) d x is

Correct Answer is : esin⁡−1x⋅x+ce^{ ^{-1} x} · x+cesin−1x⋅x+c, where c is constant of integration.

e^{ ^{-1} x}+c^{}, where c is constant of integration.

e^{ ^{-1} x} · x+c^{} where c is constant of integration.

e^{ ^{-1} x} · x+c, where c is constant of integration.

Correct Answer is : esin⁡−1x⋅x+ce^{ ^{-1} x} · x+cesin−1x⋅x+c, where c is constant of integration.

Test Index

MHTCET 11 May 2024 Shift 1 Paper

Question : 148 of 150

Marks: +1, -0

x \in[-1,1]

\int e^{\sin ^{-1} x}\left(\frac{x+\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right) d x

Solution:

Concept:

Use substitution

\sin^{-1} x = t

, then simplify the integrand to a known form

\int e^t(\sin t + \cos t)\,dt

Explanation:

Let

\sin^{-1} x = t

Then

\frac{1}{\sqrt{1-x^2}}\,dx = dt

Also,

x = \sin t

and

\sqrt{1-x^2} = \cos t

Substitute these into the integral:

I = \int e^t \left(\frac{\sin t + \cos t}{\cos t}\right)\cos t\,dt

I = \int e^t(\sin t + \cos t)\,dt

We know

\int e^t(\sin t + \cos t)\,dt = e^t\sin t + c

Replace

t

back:

e^{\sin^{-1}x} \cdot \sin(\sin^{-1}x) = e^{\sin^{-1}x} \cdot x

Thus,

I = x e^{\sin^{-1}x} + c

Answer:

Option D:

e^{\sin^{-1} x} \cdot x + c

, where

c

is the constant of integration.

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