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I'd like your help with computing the following integral:$$ \int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx.$$ I used $t=\sin x$ and got that $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^2x\cos x}{\sin x+\cos x}dx=\int_{0}^{1}\frac{t^2}{t+\sqrt{1-t^2}}dt,$$ but I can't see how I it use either. I tried also to use $t=2\arctan x$, without any successes.

What do you think?

Thank you very much!

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4 Answers 4

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The light: Denote $\displaystyle I = \int^{\pi/2}_0 \frac{ \sin^2 x \cos x}{\sin x + \cos x} dx .$ Let $ u= \pi/2 -x $ to find $\displaystyle I = \int^{\pi/2}_0 \frac{\cos^2 x \sin x}{\sin x+\cos x} dx .$ Adding these two gives $\displaystyle 2I = \int^{\pi/2}_0 \frac{ \sin x \cos x (\sin x+ \cos x) }{\sin x + \cos x} dx = \int^{\pi/2}_0 \sin x \cos x dx = \frac{\sin^2 x}{2} \biggr|^{\pi/2}_0 = \frac{1}{2}$ so $ I = \frac{1}{4}.$


The dark path: Applying the identity $ \sin x + \cos x = \sqrt{2} \sin ( x +\pi/4) $ gives $\displaystyle I = \frac{1}{\sqrt{2}} \int^{\pi/2}_0 \frac{\sin^2 x \cos x}{\sin (x+\pi/4)} dx = \frac{1}{\sqrt{2}} \int^{3\pi/4}_{\pi/4} \frac{\sin^2 (u-\pi/4) \cos (u-\pi/4)}{\sin u} du.$ Now apply $\displaystyle \sin^2 t = \frac{1-\cos 2t}{2}$ and the addition theorems, the result will eventually come.

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  • $\begingroup$ Thank you. I don't understand the light version. what did you exactly do with $u$? $\endgroup$
    – Jozef
    Commented Feb 7, 2012 at 11:05
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    $\begingroup$ Right after we let $u= \pi/2 - x$ the integral becomes $\displaystyle I = \int^{\pi/2}_0 \frac{\cos^2 u \sin u}{\sin u+\cos u} du.$ Then I just changed the letter from 'u' back to 'x', as it doesn't make a difference to the value of the integral. $\endgroup$ Commented Feb 7, 2012 at 11:18
  • $\begingroup$ Wow, I get it now. Nice!! Thanks a lot. $\endgroup$
    – Jozef
    Commented Feb 7, 2012 at 11:29
  • $\begingroup$ One second, why the both integrals (the one with $\sin^2 x$, and the other with $\cos^2 x$) are equal and both I? $\endgroup$
    – Jozef
    Commented Feb 7, 2012 at 11:44
  • $\begingroup$ If this is what's causing the confusion: $sin(x) = cos(\pi/2 - x)$ and $cos(x) = sin(\pi/2 - x)$ so after the substitution the result may look somewhat counter-intuitive. $\endgroup$
    – Sp3000
    Commented Feb 7, 2012 at 11:54
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Hint:

$$\int_{0}^{a}f(x) = \int_{0}^{a}f(a-x)$$

(Try proving this property if you don't already know it :) )

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  • $\begingroup$ +1 Surely this is the way of doing this. It is a definite integral fer cryin' out loud! $\endgroup$ Commented Feb 8, 2012 at 10:15
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I know an answer's already been accepted, but I figured I'd chime in with an answer that doesn't require nice bounds and isn't quite as ugly as Ragib's dark path. To get a more workable denominator, multiply numerator and denominator by $\cos x-\sin x$.

$\int \frac{\sin^2x\cos^2x-\sin^3x\cos x}{\cos^2x-sin^2x}dx=\int\frac{\frac14\sin^22x-\frac12\sin2x(1-\cos^2x)}{\cos2x}dx=$

$\int \frac{\frac14-\frac14\cos^22x-\frac12\sin2x(1-\frac{1+\cos2x}2)}{\cos2x}dx=\int\frac{\frac14-\frac14\cos^22x-\frac14\sin2x+\frac14\sin2x\cos2x}{\cos2x}dx=$

$\int \frac14\sec2x-\frac14\cos2x+\frac18(\frac{-2\sin2x}{\cos2x})+\frac14\sin2xdx=$

$\frac18\ln|\sec2x+\tan2x|-\frac18\sin2x+\frac18\ln|\cos2x|-\frac18\cos2x+C=$

$\frac18\ln|1+\sin2x|-\frac18\sin2x-\frac18\cos2x+C$

If you plug in the bounds given for the problem, you'll find all the sine terms equal $0$, leaving $-\frac18(\cos\pi-\cos0)=-\frac18(-1-1)=\frac14$, matching Ragib's answer.

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  • $\begingroup$ Nice! Thanks a lot! :) $\endgroup$
    – Jozef
    Commented Feb 8, 2012 at 9:36
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Then continue to follow your steps: $$I=\int_0^1\frac{t^2}{t+\sqrt{1-t^2}}dt=(let s=t^2)\int_0^1\frac{s}{\sqrt{s}+\sqrt{1-s}}\frac{ds}{2\sqrt{s}}=\frac{1}{2}\int_0^1\frac{\sqrt{s}}{\sqrt{s}+\sqrt{1-s}}ds$$ $$=\frac{1}{2}\int_0^1\frac{\sqrt{1-s}}{\sqrt{1-s}+\sqrt{s}}ds=\frac{1}{4}\int_0^1\frac{\sqrt{s}+\sqrt{1-s}}{\sqrt{1-s}+\sqrt{s}}ds=\frac{1}{4}\int_0^1ds=\frac{1}{4}$$

I think it's a little bit simple.

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