How can I handle the integation which contains $~ \left( 1+ t ^{2} \right) ^{2} ~$ and $~ \left( 1- t ^{2} \right) ^{2} ~~$?

Question

$$ \left( a,d \in \mathbb R_{> 0} \right) ~~\wedge~~ \left( a < d \right) $$

I have to find out the value of the below integral .

$$ \alpha := \int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2} \right) ^{2} }{ \left( 1+ t ^{2} \right) \left\{ d ^{2} \left( 1+ t ^{2} \right) ^{2} - a ^{2} \left( 1- t ^{2} \right) ^{2} \right\} } \,dt $$

$$ = 2a \int_{0 }^{1 } \frac{ \left( 1- t ^{2} \right) ^{2} }{ \left( 1+ t ^{2} \right) \left\{ d ^{2} \left( 1+ t ^{2} \right) ^{2} - a ^{2} \left( 1- t ^{2} \right) ^{2} \right\} } \,dt $$

My tries so far are as below .

$$ t= \tan^{}\left( x \right) $$

$$ t : 0 ~\rightarrow~1 $$

$$ x : 0 ~\rightarrow~ \frac{\pi}{4} $$

$$ \frac{ dt }{ dx } = \sec^{2}\left( x \right) $$

$$ 1+ t ^{2} = \sec^{2}\left( x\right) $$

$$ \therefore ~~ \frac{ dt }{ dx } = \left( 1 + t ^{2} \right) $$

$$ \frac{ dt }{ \left( 1+ t ^{2} \right) }= dx $$

$$ 1+ t ^{2} = \sec^{2}\left( x \right) ~~ \leftarrow~~ \text{I will subtract each right and the left term by }~~ -2 t ^{2} $$

$$ 1+ t ^{2} -2 ^{2} = \sec^{2}\left( x \right) - 2 t ^{2} $$

$$ 1- t ^{2} = \sec^{2}\left( x \right) - 2 \left( \sec^{2}\left( x \right) -1 \right) $$

$$ = \sec^{2}\left( x \right) -2 \sec^{2}\left( x \right) + 2 $$

$$ = - \sec^{2}\left( x \right) +2 $$

$$ = 2- \sec^{2}\left( x \right) $$

$$ \therefore ~~ \alpha = 2a \int_{0 }^{\frac{\pi}{4} } \frac{ \left( 2- \sec^{2}\left( x \right) \right) ^{2} }{ \left\{ d ^{2} \left( \sec^{2}\left( x \right) \right) ^{2} -a ^{2} \left( 2 - \sec^{2}\left( x_{} \right) \right)^{2} \right\} } \,dt $$

I've been got stucked from here .

Answer 1

Substitute $t=\tan \frac x2$

\begin{align} &\int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2} \right) ^{2} }{ \left( 1+ t ^{2} \right) \left\{ d ^{2} \left( 1+ t ^{2} \right) ^{2} - a ^{2} \left( 1- t ^{2} \right) ^{2} \right\} } \,dt \\ =& \int_0^{\frac\pi2}\frac{a \cos^2x}{d^2 - a^2\cos^2x}dx = \frac1a\left(-\frac\pi2 + \int_0^{\frac\pi2}\frac{d^2}{d^2 - a^2\cos^2x}dx\right)\\ = &\frac1a\left(-\frac\pi2 + \int_0^{\frac\pi2}\frac{\>d(\tan x)}{\frac{d^2 - a^2 }{d^2}+ \tan^2x}\right) =\frac\pi{2a}\left( \frac d{\sqrt{d^2-a^2}}-1\right) \end{align}