1

Show that

$$\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$$

My Attempt:

Let $$I=\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$ Using $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x)dx$ we can write:

$$I=\int_{0}^{\pi} \frac {(\pi - x)dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} $$ $$I=\int_{0}^{\pi} \frac {\pi dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - \int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}$$ $$I=2\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - I$$ $$I=\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$

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pi-π
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  • Please articulate what your question is? – amWhy Apr 01 '20 at 16:41
  • @amWhy, Proving the required. – pi-π Apr 01 '20 at 16:44
  • That's not a question. That's a command, or a statement "I was asked to prove this... here's my incomplete work". Are you implying that you are instructing us to prove it for you, or do you have a question about where you are stuck, or about the correctness of your work so far, or both? If so, *say so in an edit in your question body*, not in comments. – amWhy Apr 01 '20 at 16:47
  • Gentle suggestion: I'd recommend phrasing this in terms of "what techniques of integration are needed to perform this integral?" Have you tried trig substitutions, or rearranging the problem in any meaningful way? – Mnifldz Apr 01 '20 at 16:49

2 Answers2

4

$$I=\int_{0}^{\pi} \frac{x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(1)$$ Apply $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$. $$I=\int_{0}^{\pi} \frac{(\pi-x) dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(2)$$ Add (1) and (2) $$2I=\pi\int_{0}^{\pi} \frac{ dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(3)$$ Use $\int_{0}^{2a} f(x) dx=\int_{0}^{a} f(x) dx, ~if ~f(2a-x)=f(x)$ $$I=\pi \int_{0}^{\pi/2} \frac{dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(4)$$ Let $$J(a,b)=\int_{0}^{\pi/2} \frac{dx}{(a^2 \sin^2 x+ b^2 \cos^2x)}=\int_{0}^{\pi/2} \frac {\sec^2 x dx}{b^2+a^2\tan^2 x} =\frac{\pi}{2ab}~~~(5)$$ D. (5) w.r.t. $a$ to get $$\frac{J(a,b)}{da}=\int_{0}^{\pi/2} \frac{-2a \sin^2 xdx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=-\frac{\pi}{2a^2b}~~~~(6)$$ $$\implies \int_{0}^{\pi/2} \frac{\sin^2x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=\frac{\pi}{4a^3b}~~~(7)$$ Similarly by D.w.r.t. $b$, we can get $$\implies \int_{0}^{\pi/2} \frac{\cos^2x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=\frac{\pi}{4a^3b}~~~(8)$$ Adding (7) and (8), we get from (4) $$I=\frac{\pi^2}{4}\frac{a^2+b^2}{a^3b^2}.$$

Z Ahmed
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3

Continue with the substitution $t=\tan x$,

$$\begin{align} & \pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} \\ & = \pi \int_0^\infty \frac{1+t^2}{(b^2+a^2t^2)^2}dt \\ &=\frac{\pi(a^2-b^2)}{2a^2b^2} \frac t {b^2+a^2t^2}\bigg|_ 0^\infty + \frac{\pi(a^2+b^2)}{2a^2b^2} \int_0^\infty \frac {dt}{b^2+a^2t^2} \\ & =0+\frac{\pi(a^2+b^2)}{2a^3b^3} \tan^{-1}\frac {at}b\bigg|_0^\infty \\ &=\frac{\pi^2(a^2+b^2)}{4a^3b^3}\\ \end{align}$$

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