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We need to evaluate $\displaystyle \int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx$ and some solution to this starts as,

$\displaystyle\int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx = \int_0^{\pi/2} {\{\sin(\pi/2 -x)\}^2 \over 1 + \sin (\pi/2 -x)\cos (\pi/2 -x)}dx$.

We fail to understand how this step has been carried out. Even variable substitution does not seem to work.

Do you think that you could give us a hint?

StubbornAtom
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Masroor
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    Change of variable (substitution) $x=\pi/2-t$, and then renaming the dummy variable of integration. – André Nicolas Sep 27 '14 at 06:54
  • Geometrically, note that the graph of $\cos^2 x$ on $[0,\pi/2]$ is $\sin x$ backwards, so the area under $y=\frac{\cos^2 x}{1+\sin x\cos x}$ is the same as the area under $\frac{\sin^2 x}{1+\sin x\cos x}$. – André Nicolas Sep 27 '14 at 07:04
  • Use $\int_b^a{f(x)}dx=\int_b^a{f(a+b-x)}dx$. Then add the both to get $I= \int_0^{\pi/2}{\frac{1}{2+2sinxcosx}dx$. – Rohinb97 Sep 27 '14 at 07:19

4 Answers4

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As an alternate solution : why not to start using $\cos(2x)=1-2\sin^2(x)$ and $\sin(2x)=2 \sin(x)\cos(x)$. So, $$I=\int {\sin^2x \over 1 + \sin x\cos x}dx=\int \frac{1-\cos(2x)}{2+\sin(2x)}dx$$ Now, use the tangent half-angle substitution $t=\tan(x)$ to get $$I=\int\frac{4 t^2}{t^4+t^3+2 t^2+t+1}dt$$ Using partial fraction decomposition $$\frac{4 t^2}{t^4+t^3+2 t^2+t+1}=\frac{4 t}{t^2+1}-\frac{4 t}{t^2+t+1}$$ noticing that $$\frac{2 t}{t^2+t+1}=\frac{2 t+1-1}{t^2+t+1}=\frac{2 t+1}{t^2+t+1}-\frac{1}{t^2+t+1}$$

I am sure that you can take from here.

Claude Leibovici
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  • Thanks. Actually this was for my son. He got this problem in some practice test this morning. He attempted it up to your first step converting the angles to 2x, failing to take it further. Then the solution sheet was handed over to them after the test, which did not contain the necessary explanations. We like your solution since it fits better intuitively. – Masroor Sep 27 '14 at 07:39
  • You are very welcome ! – Claude Leibovici Sep 27 '14 at 07:42
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    @ClaudeLeibovici As an alternative to your alternative, I propose $I=\int {\sin^2x \over 1 + \sin x\cos x}dx=\frac12\int \frac{1-\cos(z)}{2+\sin(z)}dz=\frac12\int \frac{dz}{2+\sin(z)}-\frac12\int \frac{\cos(z)}{2+\sin(z)}dz=\frac12\int \frac{dt}{t^2+t+1}-\frac12\int \frac{du}{2+u}$. :) – David H Sep 27 '14 at 07:44
  • @DavidH. For sure, you are right ! In mathematics house are many mansions (as John 14:2 almost says) !! Cheers :-) – Claude Leibovici Sep 27 '14 at 07:48
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Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx,$

$\displaystyle I=\int_0^\frac\pi2\frac{\sin^2x}{1+\sin x\cos x}dx=\int_0^\frac\pi2\frac{\cos^2x}{1+\sin x\cos x}dx$

$\displaystyle2I=\int_0^\frac\pi2\frac1{1+\sin x\cos x}dx=\int_0^\frac\pi2\frac{\sec^2x}{1+\tan^2x+\tan x}dx$

Setting $\tan x=u$

$\displaystyle2I=\int_0^\infty\frac{dt}{1+t+t^2}=4\int_0^\infty\frac{dt}{(2t+1)^2+(\sqrt3)^2}$

Set $2t+1=\sqrt3\tan\theta$

lab bhattacharjee
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Using double-angle formula, we get $$I=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2+\sin 2 x} d x$$ $$I\stackrel{x\mapsto\frac{\pi}{4}-x}{=} 2 \int_{0}^{\frac{\pi}{4}} \frac{1}{2+\cos 2 x} d x-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sin 2 x}{2+\cos 2 x} d x$$ Since $\dfrac{1}{2+\cos 2 x}$ is even and $\dfrac{\sin 2 x}{2+\cos 2 x}$ is odd, therefore

$$ \displaystyle I=2 \int_{0}^{\frac{\pi}{4}} \frac{1}{1+2 \cos ^{2} x} d x$$ $$\displaystyle I=2 \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{2} x}{\sec ^{2} x+2} d x \displaystyle \stackrel{t=\tan x}{=} 2 \int_{0}^{1} \frac{d t}{3+t^{2}}=\dfrac{\pi}{3 \sqrt{3}} $$

Lai
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I think expand $\sin(\frac{\pi}{2} -2x)$ and $\cos(\frac{\pi}{2} -2x)$ by using sine rule and cosine rule will make it clearer and easier to integrate.

J.G.
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Wang Kah Lun
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  • By sine and cosine rule, do you mean compound angle formulae? Usually, those terms refer to identifies for triangles. – J.G. Mar 09 '23 at 13:36