If A=∫0π cos⁡x(x+2)2dx then ∫0π/2 sin⁡2x(x+1)dx is equal to

A=∫0π cos⁡x(x+2)2dx=∫0π (cos⁡x)1(x+2)2dx By using by parts =cos⁡x−1x+20π−∫0π (−sin⁡x)−1x+2dx =--1π+2-12-∫0πsinxx+2dx put x=2y differentiate w.r.to 'y' on both sides then dx=2dy If x=0⇒y=0 and x=π⇒y=π2 then above integral becomes A=1π+2+12-∫0π2 sin⁡2y2y+22dy A=1π+2+12-∫0π2 sin⁡2yy+1dy ∴A=1π+2+12−∫0π/2 sin⁡2xx+1dx ⇒∫0π/2 sin⁡2xx+1dx=1π+2+12-A

If A=∫0π cos⁡x(x+2)2dx then ∫0π/2 sin⁡2x(x+1)dx is equal to

A=∫0π cos⁡x(x+2)2dx=∫0π (cos⁡x)1(x+2)2dx By using by parts =cos⁡x−1x+20π−∫0π (−sin⁡x)−1x+2dx =--1π+2-12-∫0πsinxx+2dx put x=2y differentiate w.r.to 'y' on both sides then dx=2dy If x=0⇒y=0 and x=π⇒y=π2 then above integral becomes A=1π+2+12-∫0π2 sin⁡2y2y+22dy A=1π+2+12-∫0π2 sin⁡2yy+1dy ∴A=1π+2+12−∫0π/2 sin⁡2xx+1dx ⇒∫0π/2 sin⁡2xx+1dx=1π+2+12-A

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If $A = \int_{0}^{π} \frac{\cos x}{(x + 2)^{2}} dx then \int_{0}^{π / 2} \frac{\sin 2 x}{(x + 1)} dx$ is equal to

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$A - \frac{1}{2} - \frac{1}{π + 2}$

$\frac{1}{2} + \frac{1}{π + 2} - A$

$\frac{1}{π + 2} - A$

$1 + \frac{1}{π + 2} - A$

answer is B.

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Detailed Solution

$A = \int_{0}^{π} \frac{\cos x}{(x + 2)^{2}} dx = \int_{0}^{π} (\cos x) \frac{1}{(x + 2)^{2}} dx B y u \sin g b y p a r t s = {[\cos x (- \frac{1}{x + 2})]}_{0}^{π} - \int_{0}^{π} (- \sin x) (- \frac{1}{x + 2}) dx = - [\frac{- 1}{π + 2} - \frac{1}{2}] - \int_{0}^{π} \frac{\sin x}{x + 2} d x p u t x = 2 y d i f f e r e n t i a t e w . r . t o' y' o n b o t h s i d e s t h e n d x = 2 d y I f x = 0 \Rightarrow y = 0 a n d x = π \Rightarrow y = \frac{π}{2} t h e n a b o v e i n t e g r a l b e c o m e s A = [\frac{1}{π + 2} + \frac{1}{2}] - \int_{0}^{\frac{π}{2}} \frac{\sin 2 y}{2 y + 2} 2 dy A = [\frac{1}{π + 2} + \frac{1}{2}] - \int_{0}^{\frac{π}{2}} \frac{\sin 2 y}{y + 1} dy ∴ A = \frac{1}{π + 2} + \frac{1}{2} - \int_{0}^{π / 2} \frac{\sin 2 x}{x + 1} dx \Rightarrow \int_{0}^{π / 2} \frac{\sin 2 x}{x + 1} dx = \frac{1}{π + 2} + \frac{1}{2} - A$