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Link ∫cosx/(x+2)² and ∫sin2x/(x+1) | JEE

JEE Maths question with a full step-by-step solution.

Question

A=\displaystyle\int_{0}^{\pi}\dfrac{\cos x}{(x+2)^{2}}\,dx

, then

\displaystyle\int_{0}^{\pi/2}\dfrac{\sin 2x}{x+1}\,dx

is equal to

A-\dfrac12-\dfrac{1}{\pi+2}

\dfrac12+\dfrac{1}{\pi+2}-A

correct

\dfrac{1}{\pi+2}-A

1+\dfrac{1}{\pi+2}-A

Solution

Step 1: Integrate

A

by parts, taking

u=\cos x

and

dv=\dfrac{dx}{(x+2)^{2}}

so that

v=-\dfrac{1}{x+2}

A=\left[-\dfrac{\cos x}{x+2}\right]_{0}^{\pi}-\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx.

The boundary term is

-\dfrac{\cos\pi}{\pi+2}+\dfrac{\cos 0}{2}=\dfrac{1}{\pi+2}+\dfrac12

, hence

A=\dfrac{1}{\pi+2}+\dfrac12-\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx.

Step 2: Transform the remaining integral. In

\displaystyle\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx

put

x=2t

dx=2\,dt

, with

t:0\to\dfrac{\pi}{2}

\int_{0}^{\pi}\dfrac{\sin x}{x+2}\,dx=\int_{0}^{\pi/2}\dfrac{\sin 2t}{2t+2}\cdot 2\,dt=\int_{0}^{\pi/2}\dfrac{\sin 2t}{t+1}\,dt.

Step 3: Substitute back into the expression for

A

A=\dfrac{1}{\pi+2}+\dfrac12-\int_{0}^{\pi/2}\dfrac{\sin 2t}{t+1}\,dt.

Step 4: Solve for the required integral:

\int_{0}^{\pi/2}\dfrac{\sin 2x}{x+1}\,dx=\dfrac12+\dfrac{1}{\pi+2}-A.

Correct answer: (2)

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