Given that, $$ I=\displaystyle\int\limits_{0}^{\frac{\pi}{2}} \dfrac{\mathrm{d}x}{a^2\sin^2 x+b^2\cos^2 x}=\dfrac{\pi}{2ab}$$ Evaluate: $$ J=\displaystyle\int\limits_{0}^{\frac{\pi}{2}} \dfrac{\mathrm{d}x}{(a^2\sin^2 x+b^2\cos^2 x)^2}$$

MY WORK: Using differentiating under the integral sign with respect to $a^2$ and $b^2$, I obtained the equations, $$\displaystyle\int\limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x \mathrm{d}x}{(a^2\sin^2 x+b^2\cos^2 x)^2}=\dfrac{\pi}{4a^3b}\tag{1}$$ $$\displaystyle\int\limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x \mathrm{d}x}{(a^2\sin^2 x+b^2\cos^2 x)^2}=\dfrac{\pi}{4ab^3}\tag{2}$$

Adding $(1)$ and $(2)$ gives, $$J=\dfrac{\pi(a^2+b^2)}{4a^3b^3}$$ Is this correct?

  • $\begingroup$ $Mathematica$ gives $J=\frac{\pi \left(\frac{a^2}{b^2}\right)^{3/2} \left(a^2+b^2\right)}{4 a^6}\quad$ Differentiating under the integral wrt $a$ results, according to my software, $-\frac{4 a \sin ^2(x)}{\left(a^2 \sin ^2(x)+b^2 \cos ^2(x)\right)^3}$ $\endgroup$ – Raffaele Aug 5 '17 at 9:18
  • $\begingroup$ You misunderstood me. I differentiated under the integral with respect to $a^$$. $\endgroup$ – Archimedesprinciple Aug 5 '17 at 9:27
  • $\begingroup$ It is correct. Why do you doubt your procedure? $\endgroup$ – Paramanand Singh Aug 5 '17 at 11:12

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