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Show that $\int_{0}^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx=\frac{1}{2}(\frac{1}{2}+\frac{1}{\pi +2}-A)$ where $A=\int_0^\pi\frac{\cos x}{(x+2)^2}dx$.

I tried using partial integration on the integral $\int_0^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx$ but I am in no luck. Also tried using $\frac12A=\int_0^{\pi/2}\frac{\cos x}{(x+2)^2}dx$. But still something is going wrong . Please suggest.

The best I could think of is :

$\int {\dfrac{sin2x}{2x+2}dx}=-\dfrac{1}{2x+2}cos2x+\dfrac{1}{4}\int {\dfrac{cos2x}{(x+1)^2}dx}$

Notice carefully that $\int_0^\pi\frac{\cos x}{(x+2)^2}dx$ has the same structure as $\int {\dfrac{cos2x}{(x+1)^2}dx}$

P.S. See I purposefully avoided the long details of my efforts as that would hamper the understandibility of the problem and may mislead the answerer as well.

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  • $\begingroup$ Are the denominators different? Or is that a typo? $\endgroup$
    – N. S.
    Commented Apr 27, 2018 at 19:36
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    $\begingroup$ @N.S. No I also first thought the same. It is not! I am just scribbling and scribbling pages trying to figure that out! I am sorry !Sometimes I start with the integral to end at the same place!Like a labyrinth! $\endgroup$
    – Saradamani
    Commented Apr 27, 2018 at 19:37
  • $\begingroup$ Don't know if $\sin 2x = 2 \sin x \cos x$ will help... $\endgroup$
    – Alex Vong
    Commented Apr 27, 2018 at 19:40
  • $\begingroup$ @AlexVong No I tried it but I atleast got no result as of now! $\endgroup$
    – Saradamani
    Commented Apr 27, 2018 at 19:40

3 Answers 3

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Just a couple of tiny missteps by the OP.

\begin{align*} \mathcal{I} \equiv\int_0^{\pi/2} \frac{ \sin x \cos x}{x+1}\,\mathrm{d}x &= \int_0^{\pi/2} \frac{ \sin 2x }{2x+2}\,\mathrm{d}x \tag*{, then denote $u = 2x$} \\ &= \frac12\int_0^{\pi} \frac{ \color{blue}{\sin u} }{u+2}\,\color{blue}{\mathrm{d}u} \tag*{, then by-part ...} \end{align*} ... let $\color{blue}{\sin u \,\mathrm{d}u}$ go together \begin{align*} \mathcal{I} &= \frac12 \left[ \frac{ -\cos u }{u+2}\Bigg|_0^{\pi} - \int_0^{\pi} (-\cos u) \cdot \frac{ -1 }{ (u+2)^2 }\,\mathrm{d}u\right ] \\ &= \frac12 \left[ \frac1{ \pi + 2} - \frac{-1}{2} - \int_0^{\pi} \frac{ \cos u}{ (u+2)^2 }\,\mathrm{d}u\right] \end{align*} which is the desired result.

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You are on the right lines!

$$\small\int_0^{\pi/2} {\frac{\sin2x}{2x+2}dx}=\left[-\frac{1}{2x+2}\cos2x\right]_0^{\pi/2}-\int_0^{\pi/2} {\frac{\cos2x}{(2x+2)^2}dx}=\frac12\left(\frac12+\frac1{\pi+2}\right)-\int_0^{\pi/2} {\frac{\cos2x}{(2x+2)^2}dx}$$ Now use the substitution $u=2x$.

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The change of limits from $\frac{\pi}{2}$ to $\pi$ suggests the substitution $u = 2x$.

This then suggests that you need to rewrite $\sin(x)\cos(x)$ in terms of trigonometric functions of $2x$. So the next step is to consult double angle formulae.

You've already figured out that you'll need partial integration, so you know you have something that either differentiates or integrates to $\cos$.

So what function of $2x$ has a derivative or integral of $\cos$, and is equal to a product of $\sin$ and $\cos$?

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    $\begingroup$ This is better suited as a comment. $\endgroup$
    – TheSimpliFire
    Commented Apr 27, 2018 at 19:56

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