ISI (2)

Calculus Level 3

Given that A = ∫ 0 π cos ⁡ x ( x + 2 ) 2 d x \displaystyle A=\int_0^\pi \frac{\cos x}{(x+2)^2} dx , what is the value of ∫ 0 π 2 sin ⁡ x cos ⁡ x x + 1 d x ? \displaystyle \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{x+1} dx?

1 2 ( 1 2 + 1 π + 2 − 1 A ) \frac{1}{2}\left(\frac{1}{2}+\frac{1}{\pi+2}-\frac{1}{A}\right) 1 2 ( 1 4 + 1 π + 2 − A 2 ) \frac{1}{2}\left(\frac{1}{4}+\frac{1}{\pi+2}-A^2\right) 1 4 ( 1 2 + 1 2 π + 2 − A ) \frac{1}{4}\left(\frac{1}{2}+\frac{1}{2\pi+2}-A\right) 1 4 ( 1 4 + 1 π + 2 − A ) \frac{1}{4}\left(\frac{1}{4}+\frac{1}{\pi+2}-A\right) 1 2 ( 1 2 + 1 π + 2 − A ) \frac{1}{2}\left(\frac{1}{2}+\frac{1}{\pi+2}-A\right)

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
May 15, 2018

Relevant wiki: Integration by Parts - Easy

Consider

I = ∫ 0 π 2 sin ⁡ x cos ⁡ x x + 1 d x = 1 2 ∫ 0 π 2 sin ⁡ ( 2 x ) x + 1 d x Let u = 2 x ⟹ d u = 2 d x = 1 2 ∫ 0 π sin ⁡ u u + 2 d u \begin{aligned} I & = \int_0^\frac \pi 2 \frac {\sin x \cos x}{x+1} dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac {\sin (2x)}{x+1} dx & \small \color{#3D99F6} \text{Let }u = 2x \implies du = 2\ dx \\ & = \frac 12 \int_0^\pi \frac {\sin u}{u+2} du \end{aligned}

Now,

A = ∫ 0 π cos ⁡ x ( x + 2 ) 2 d x By integration by parts = − cos ⁡ x x + 2 ∣ 0 π − ∫ 0 π sin ⁡ x x + 2 d x = 1 π + 2 + 1 2 − 2 I \begin{aligned} A & = \int_0^\pi \frac {\cos x}{(x+2)^2}dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = - \frac {\cos x}{x+2} \bigg|_0^\pi -\color{#3D99F6} \int_0^\pi \frac {\sin x}{x+2}dx \\ & = \frac 1{\pi +2} + \frac 12 - \color{#3D99F6}2I \end{aligned}

Therefore, I = 1 2 ( 1 2 + 1 π + 2 − A ) I = \boxed{\dfrac 12 \left(\dfrac 12 + \dfrac 1{\pi + 2} - A\right)} .

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