For 0 < a < 1 , the value of the integral ∫ _ 0 ^ π dx 1 - 2acosx +...

Given: $I = {∫}_{0}^{\pi}\dfrac{dx}{1 - 2acosx + {a}^{2}} . . . \left(i\right)$

$⇒ I = {∫}_{0}^{\pi}\dfrac{dx}{1 - 2acos\left(\pi - x\right) + {a}^{2}}$

$⇒ I = {∫}_{0}^{\pi}\dfrac{dx}{1 + 2acosx + {a}^{2}} . . . \left(ii\right)$

Adding \left(i\right)$and \left(ii\right)$,

$⇒ 2I = {∫}_{0}^{\pi}\dfrac{dx}{1 - 2acosx + {a}^{2}} + {∫}_{0}^{\pi}\dfrac{dx}{1 + 2acosx + {a}^{2}}$

$⇒ 2I = {∫}_{0}^{\pi}\left(\dfrac{1}{1 - 2acosx + {a}^{2}} + \dfrac{1}{1 + 2acosx + {a}^{2}}\right)dx$

$⇒ 2I = {∫}_{0}^{\pi}\left(\dfrac{1 + 2acosx + {a}^{2} + 1 - 2acosx + {a}^{2}}{{\left(1 + {a}^{2}\right)}^{2} - {\left(2acosx\right)}^{2}}\right)dx$

$⇒ 2I = {∫}_{0}^{\pi}\left(\dfrac{2\left(1 + {a}^{2}\right)}{{\left(1 + {a}^{2}\right)}^{2} - {\left(2acosx\right)}^{2}}\right)dx$

$⇒ 2I = 2{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{2\left(1 + {a}^{2}\right)}{{\left(1 + {a}^{2}\right)}^{2} - 4{a}^{2}\cos^{2}x}dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{2\left(1 + {a}^{2}\right) \cdot \sec^{2}x}{{\left(1 + {a}^{2}\right)}^{2} \cdot \sec^{2}x - 4{a}^{2}}dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{2\left(1 + {a}^{2}\right) \cdot \sec^{2}x}{{\left(1 + {a}^{2}\right)}^{2} + {\left(1 + {a}^{2}\right)}^{2}\tan^{2}x - 4{a}^{2}}dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{2\left(1 + {a}^{2}\right) \cdot \sec^{2}x}{{\left(1 + {a}^{2} - 2a\right)}^{2} + {\left(1 + {a}^{2}\right)}^{2}\tan^{2}x}dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{2 \cdot \left(1 + {a}^{2}\right) \cdot \sec^{2}x}{{\left(1 + {a}^{2}\right)}^{2} \cdot \tan^{2}x + {\left(1 - {a}^{2}\right)}^{2}}dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\dfrac{2 \cdot \sec^{2}x}{1 + {a}^{2}} \cdot dx}{\tan^{2}x + {\left(\dfrac{1 - {a}^{2}}{1 + {a}^{2}}\right)}^{2}}$

Let, $\tan x = t$

$⇒ \sec^{2}xdx = dt$

$⇒ I = \dfrac{2}{1 + {a}^{2}}{∫}_{0}^{\infty}\dfrac{dt}{{t}^{2} + {\left(\dfrac{1 - {a}^{2}}{1 + {a}^{2}}\right)}^{2}}$

$⇒ I = \dfrac{2}{1 + {a}^{2}} \times \dfrac{1}{\left(\dfrac{1 - {a}^{2}}{1 + {a}^{2}}\right)}{\left[\tan^{- 1}\dfrac{t}{\left(\dfrac{1 - {a}^{2}}{1 + {a}^{2}}\right)}\right]}_{0}^{\infty}$

$⇒ I = \dfrac{2}{1 - {a}^{2}}\left[\tan^{- 1}\left(\infty\right) - \tan^{- 1}\left(0\right)\right]$

$⇒ I = \dfrac{2}{\left(1 - {a}^{2}\right)}\left[\dfrac{\pi}{2} - 0\right]$

$⇒ I = \dfrac{\pi}{1 - {a}^{2}}$