{A} = ₀^{π} x \, dx/ x + x { and } {B} = ₀^{π} x \, dx/ x - x What is the value of B?

Correct Answer is : π2{{{π }}}{2}2π

Correct Answer is : π2{{{π }}}{2}2π

Test Index

UPSC NDA II 2015 Mathematics Paper

Question Numbers: 81-82

For the next two (02) items that follow:

Consider the integrals

Question : 82 of 120

Marks: +1, -0

\mathrm{A} = \int_{0}^{\pi} \frac{\sin x \, dx}{\sin x + \cos x} \text{ and } \mathrm{B} = \int_{0}^{\pi} \frac{\sin x \, dx}{\sin x - \cos x}

What is the value of B?

Solution:

Concept:

Use the property

\int_0^a f(x) dx = \int_0^a f(a-x) dx

and symmetry to simplify the integral.

Explanation:

Let

B = \int_0^{\pi} \frac{\sin x}{\sin x - \cos x} \, dx

Apply the property:

B = \int_0^{\pi} \frac{\sin(\pi - x)}{\sin(\pi - x) - \cos(\pi - x)} \, dx

Since

\sin(\pi - x) = \sin x

and

\cos(\pi - x) = -\cos x

, we get

B = \int_0^{\pi} \frac{\sin x}{\sin x + \cos x} \, dx

Define

I = \int_0^{\pi} \frac{\sin x}{\sin x + \cos x} \, dx

and

J = \int_0^{\pi} \frac{\cos x}{\sin x + \cos x} \, dx

Then

I + J = \int_0^{\pi} 1 \, dx = \pi

Also

I - J = \int_0^{\pi} \frac{\sin x - \cos x}{\sin x + \cos x} \, dx

Substitute

u = \sin x + \cos x

, hence

du = (\cos x - \sin x) dx = -(\sin x - \cos x) dx

I-J = -\int_{x=0}^{\pi} \frac{du}{u} = -\left[ \ln|u| \right]_{0}^{\pi} = -\left( \ln|\sin\pi+\cos\pi| - \ln|\sin0+\cos0| \right)

x=0

\sin0+\cos0=1

; at

x=\pi

\sin\pi+\cos\pi=-1

Thus

I-J = -\left( \ln1 - \ln1 \right) = 0

, so

I = J

From

I+J=\pi

, we get

2I=\pi

, hence

I = \frac{\pi}{2}

Therefore

B = \frac{\pi}{2}

Answer:

B = \frac{\pi}{2}

(Option B).

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