I have to integral the below formula of integration .
$$ a, d \in\mathbb{R}_{>0} ~~\wedge~~ d > a $$
$$ \alpha := \int_{0 }^{\frac{\pi}{2} } \frac{ d }{ d- a \cdot \cos^{}\left(\theta_{} \right) } \,d\theta $$
$$ = d\int_{0 }^{\frac{\pi}{2} } \frac{ 1 }{ d- a \cdot \cos^{}\left(\theta_{} \right) } \,d\theta $$
My tries are as below so far . By the way , I wonder if such system exists which enable 2-column-display to be used against the below equations .
$$ t= \tan^{}\left( \frac{\theta_{}}{ 2 } \right) $$
$$ \theta_{} : 0 ~\rightarrow~ \frac{\pi}{2} $$
$$ t : 0 ~\rightarrow~1= \tan^{}\left( \frac{ 1 }{ 2 } \cdot \frac{\pi}{2} \right) $$
$$ \frac{ dt }{ d\theta } = \frac{1}{2} \sec^2\left( \frac{\theta}{ 2 } \right) $$
$$ \frac{ d\theta }{ dt } = \frac{ 2 }{ \sec^2\left( \frac{\theta}{2} \right) } $$
$$ 1+ t^2 = \sec^2\left( \frac{\theta}{2} \right) $$
$$ \therefore ~~ d\theta = \frac{ 2 \,dt }{ 1+ t^2 } = 2 \, dt \left( 1 + t^2 \right)^{-1} $$
$$ \cos(\theta) = 2 \cos^2\left(\frac{\theta}{ 2 } \right) -1 $$
$$ = 2 \left( \cos\left(\frac{\theta}{ 2 } \right) \right)^2 -1 $$
$$ = 2 \left( \sec^{-1}\left( \frac{\theta}{2} \right) \right)^2 -1 $$
$$ = 2 \left( \sec^2\left( \frac{\theta}{2} \right) \right)^{-1} -1 $$
$$ \therefore ~~ \cos(\theta) = 2(1+ t^2)^{-1} -1 $$
$$ \cos(\theta) = \frac 2 {1+ t^2} -1 $$
$$ = \frac 2 {1 + t^2} - \frac{ (1+ t^2) }{1+t^2} $$
$$ = \frac{ 2 - (1 + t^2) }{ 1+ t^2} $$
$$ = \frac{ 2-1-t^2}{1+ t^2} $$
$$ = \frac{ 1- t^2}{ 1 + t^2} = \cos(\theta) = (1-t^2) (1+t^2)^{-1} $$
$$ \alpha = \int_0^{\frac{\pi}{2} } \frac d { d- a \cos(\theta)} \,d\theta $$
$$ = d\int_{0 }^{\frac{\pi}{2} } \frac{ d\theta }{ d- a \cdot \cos^{}\left(\theta_{} \right) } $$
$$ = d\int_{0 }^{1 } \left( \frac{ 2 \, dt \left( 1 + t ^{2} \right) ^{-1} }{ d- a \cdot \left\{ \left( 1- t ^{2} \right) \left( 1+t ^{2} \right) ^{-1} \right\} } \right) $$
$$ = d \int_{0 }^{1 } \frac{ 2 \,dt (1+ t^2)^{-1} }{ d - a(1- t^2) (1+ t^2)^{-1}} $$
$$ = 2d \int_0^1 \frac{ (1+ t^2)^{-1} \,dt }{ d - a(1-t^2) (1+t^2)^{-1}} $$
$$ = 2d \int_0^1 \frac{(1+t^2)^{-1} \, dt }{ \left\{d - a(1- t^2) (1+ t^2)^{-1} \right\} } \cdot \frac{ (1 + t^2)^1 }{ (1 + t^2)^1} $$
$$ = 2 d \int_{0 }^{1 } \frac{ dt }{ d \left( 1+ t ^{2} \right) -a \cdot \left( 1-t ^{2} \right) } $$
$$ = 2 d \int_{0 }^{1 } \frac{ 1 }{ d + d \cdot t ^{2} - a + a \cdot t ^{2} } dt $$
$$ = 2 d \int_{0 }^{1 } \frac{ 1 }{ \left( d-a \right) + \left( d+a \right)t ^{2} } dt $$
$$ = 2d \left[ \frac{ \ln\left( \left( d-a \right) + \left( d+ a \right) t ^{2} \right) }{ 2 \left( d + a \right)t } \right]_{0}^{1} ~~ \leftarrow~~ \text{As } t= 0 ~~\text{is set , the zero division error occurs .} $$
I need hints to solve $~ \alpha ~$
Use the below general equation to proceed the given integration .
$$ a \in \mathbb R_{> 0} $$
$$ \frac{ a }{ a ^{2} + x ^{2} } = \frac{ d }{ dx } \left( \tan^{-1} \left( \frac{ x }{a } \right) \right) $$