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Mathematics Question on Integrals of Some Particular Functions

Suppose M=∫0π/2cos⁡xx+2dxM=\int\limits_{0}^{\pi / 2} \frac{\cos x}{x+2} d x, N=∫0π/4sin⁡xcos⁡x(x+1)2dxN=\int\limits_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} d x Then, the value of (M−N)(M-N) equals

A

3π+2\frac{3}{\pi+2}

B

2π−4\frac{2}{\pi-4}

C

4π−2\frac{4}{\pi-2}

D

2π+4\frac{2}{\pi+4}

Answer

2π+4\frac{2}{\pi+4}

Explanation

Solution

Given, M=∫0π/2cos⁡x(x+2)dxM=\int\limits_{0}^{\pi / 2} \frac{\cos x}{(x+2)} d x
and N=∫0π/4sin⁡xcos⁡x(x+1)2dxN=\int\limits_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} d x
⇒N=∫0π/412sin⁡2x(x+1)2dx\Rightarrow N=\int\limits_{0}^{\pi / 4} \frac{1}{2} \frac{\sin 2 x}{(x+1)^{2}} d x
Put2x=t2 x =t
⇒dx=dt2\Rightarrow d x=\frac{d t}{2}
∴N=∫0π/2sin⁡t4(t2+1)2dt\therefore N=\int\limits_{0}^{\pi / 2} \frac{\sin t}{4\left(\frac{t}{2}+1\right)^{2}} d t
=∫0π/2sin⁡t(t+2)2dt=\int\limits_{0}^{\pi / 2} \frac{\sin t}{(t+2)^{2}} d t
=∫0π/2sin⁡x(x+2)2dx=\int\limits_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x
∴M−N=∫0π/2cos⁡xII×1(x+2)Idx−∫0π/2sin⁡x(x+2)2dx\therefore M-N=\int\limits_{0}^{\pi / 2} \underset{II}{\cos x} \times \underset{I}{\frac{1}{(x+2)}} d x -\int\limits_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x
=[sin⁡xx+2]0π/2−∫0π/2−sin⁡x(x+2)2dx−∫0π/2sin⁡x(x+2)2dx=\left[\frac{\sin x}{x+2}\right]_{0}^{\pi / 2}-\int\limits_{0}^{\pi / 2}-\frac{\sin x}{(x+2)^{2}} d x -\int\limits_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x
=sin⁡π2π2+2=1π+42=2π+4=\frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}+2}=\frac{1}{\frac{\pi+4}{2}}=\frac{2}{\pi+4}