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UPSC NDA I 2019 Mathematics Solved Paper

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Question : 85 of 120

Marks: +1, -0

\int_{0}^{\frac{\pi}{2}} e^{\sin x} \cos x \; dx

is equal to

e + 1
e - 1
e + 2
e

Solution:

Concept:

The integral is of the form

\int e^{f(x)} f'(x) \, dx = e^{f(x)} + C

.

Explanation:

Let

t = \sin x

.

When

x = 0

,

t = 0

.

When

x = \frac{\pi}{2}

,

t = 1

.

Differentiating both sides with respect to

x

, we get

dt = \cos x \, dx

.

Substitute into the integral:

\displaystyle \int_0^{\pi/2} e^{\sin x} \cos x \, dx = \int_0^1 e^t \, dt

.

Evaluate:

\left[ e^t \right]_0^1 = e^1 - e^0 = e - 1

.

Answer:

e - 1

(Option B)

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