MHB Can You Solve This Complex Trigonometric Integral?

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The discussion focuses on solving the integral \( I(a,b) = \int_{0}^{\pi / 2}\frac{1}{(a\cdot\cos^2{x}+b\cdot\sin^2{x})^2}dx \) for positive constants \( a \) and \( b \). Participants derive the solution using differentiation under the integral sign and transformations, leading to the result \( I(a,b) = \frac{\pi}{4\sqrt{ab}}\left( \frac{1}{a}+\frac{1}{b}\right) \). One contributor also explores a related integral \( J \) and its derivatives, demonstrating a similar approach to arrive at the solution. The discussion highlights the effectiveness of differentiation techniques in evaluating complex integrals. The final result showcases the interplay between trigonometric functions and integral calculus.
sbhatnagar
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Can you solve the Integral?

\[ I(a,b)=\int_{0}^{\pi / 2}\frac{1}{(a\cdot\cos^2{x}+b\cdot\sin^2{x})^2}dx \quad a,b>0 \]
 
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Elementary, my dear Sbhatnagar. Let's first calculate:$$\displaystyle \begin{aligned} I( \lambda) & = \int_{0}^{\pi/2}\frac{1}{a~\cos^2{x}+b~\sin^2{x}-\lambda}\;{dx} \\& = \int_{0}^{\pi/2}\frac{1}{(a-\lambda)\cos^2{x}+(b-\lambda)\sin^2{x}}\;{dx} \\& = \int_{0}^{\pi/2}\frac{\sec^2{x}}{(a-\lambda)+(b-\lambda)\tan^2{x}}\;{dt} \\& = \int_{0}^{\infty}\frac{1}{(a-\lambda)+(b-\lambda)t^2}\;{dt} \\& = \frac{1}{\sqrt{a-\lambda}\sqrt{b-\lambda}}\bigg(\frac{\sqrt{b-\lambda}}{\sqrt{a-\lambda}}t\bigg)_{0}^{\infty}\\& = \frac{\pi}{2\sqrt{a-\lambda}\sqrt{b-\lambda}}.\end{aligned} $$

But $\displaystyle I(a, b) = I'(\lambda)|_{\lambda = 0} = \frac{(a+b-2\lambda)\pi}{4(a-\lambda)^{\frac{3}{2}}((b-\lambda)^{\frac{3}{2}}}\bigg|_{\lambda = 0} = \frac{(a+b)\pi}{4a^{\frac{2}{3}}b^{\frac{2}{3}}}.$

 
Nice! I also solved the problem by a similar approach.

Let \( \displaystyle J=\int_{0}^{\pi / 2} \frac{1}{a\cos^2{x}+b\sin^2{x}}dx \)

\[ \begin{align*} J &= \int_{0}^{\pi / 2} \frac{1}{a\cos^2{x}+b\sin^2{x}}dx \\ &= \int_{0}^{\infty} \frac{1}{a+bx^2}dx \\ &= \frac{1}{\sqrt{ab}} \tan^{-1}{\left(\frac{x\sqrt{b}}{\sqrt{a}} \right)}\Big|_0^{\infty}\\ &= \frac{\pi}{2\sqrt{ab}}\end{align*} \]

Now,

\( \displaystyle \begin{align*} \frac{dJ}{da}&=-\int_{0}^{\pi / 2}\frac{\cos^2(x)}{(a\cos^2{x}+b\sin^2{x})^2} dx\\ \frac{\pi}{4\sqrt{a^3 b}} &= \int_{0}^{\pi / 2}\frac{\cos^2(x)}{(a\cos^2{x}+b\sin^2{x})^2} dx \quad [1]\end{align*} \)

also

\( \displaystyle \begin{align*} \frac{dJ}{db}&=-\int_{0}^{\pi / 2}\frac{\sin^2(x)}{(a\cos^2{x}+b\sin^2{x})^2} dx\\ \frac{\pi}{4\sqrt{a b^3}} &= \int_{0}^{\pi / 2}\frac{\sin^2(x)}{(a\cos^2{x}+b\sin^2{x})^2} dx \quad [2]\end{align*} \)

Adding [1] and [2]:

\( \displaystyle \begin{align*} \int_{0}^{\pi / 2}\frac{\sin^2(x)+\cos^2(x)}{(a\cos^2{x}+b\sin^2{x})^2} dx &= \frac{\pi}{4\sqrt{a^3 b}}+\frac{\pi}{\sqrt{b^3 a}} \\ \int_{0}^{\pi / 2}\frac{1}{(a\cos^2{x}+b\sin^2{x})^2}dx &= \frac{\pi}{4\sqrt{ab}}\left( \frac{1}{a}+\frac{1}{b}\right) \\ I(a,b) &= \frac{\pi}{4\sqrt{ab}}\left( \frac{1}{a}+\frac{1}{b}\right)\end{align*}\)
 
Magic differentiation for the win! (Smile)
 
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