Can You Solve This Complex Trigonometric Integral?

[sbhatnagar]

Can you solve the Integral?

\[ I(a,b)=\int_{0}^{\pi / 2}\frac{1}{(a\cdot\cos^2{x}+b\cdot\sin^2{x})^2}dx \quad a,b>0 \]

 

[Sherlock1]

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

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[sbhatnagar]

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]:

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

 

[Sherlock1]

Magic differentiation for the win! (Smile)

 

[CellCoree]

Yes, I can solve this integral. First, we can rewrite the integral as:

\[ I(a,b)=\int_{0}^{\pi / 2}\frac{1}{(a+b)\cdot\cos^2{x}+(a-b)\cdot\sin^2{x})^2}dx \]

Next, we can use the trigonometric identity $\cos^2{x}+\sin^2{x}=1$ to simplify the denominator:

\[ I(a,b)=\int_{0}^{\pi / 2}\frac{1}{(a+b)+(a-b)\cdot\sin^2{x})^2}dx \]

Now, we can use the substitution $u=\sin{x}$ to transform the integral into a form that can be solved using partial fractions:

\[ I(a,b)=\int_{0}^{1}\frac{1}{(a+b)+(a-b)u^2)^2}du \]

Using partial fractions, we can rewrite the integrand as:

\[ \frac{1}{(a+b)+(a-b)u^2)^2}=\frac{1}{4ab}\left(\frac{1}{a+b+(a-b)u}+\frac{1}{a+b-(a-b)u}\right)-\frac{1}{4ab(a+b)^2} \]

Substituting this back into the integral and solving, we get:

\[ I(a,b)=\frac{1}{4ab}\left[\frac{\arctan{\left(\frac{a+b+(a-b)}{2\sqrt{ab}}\right)}-\arctan{\left(\frac{a+b-(a-b)}{2\sqrt{ab}}\right)}}{\sqrt{a+b}}\right]_{0}^{1}-\frac{1}{4ab(a+b)^2} \]

Simplifying, we get:

\[ I(a,b)=\frac{\pi}{8ab\sqrt{a+b}}-\frac{1}{4ab(a+b)^2} \]

Therefore, the solution to the integral is:

\[ I(a,b)=\frac{\pi}{8ab\sqrt{a+b}}-\frac{1}{4ab(a+b)^2} \]