For 0<a<1, the value of the integral ∫ _0^π dx 1-2a cosx+a^2 is :

Given: $\displaystyle I={\int }_{0}^{\pi }\frac{\mathrm{dx}}{1-2\mathrm{acosx}+{a}^{2}}...(i)$I=∫0π​1−2acosx+a2dx​...(i)

$$\Rightarrow I={\int }_{0}^{\pi }\frac{\mathrm{dx}}{1-2\mathrm{acos}(\pi -x)+{a}^{2}}$$⇒I=∫0π​1−2acos(π−x)+a2dx​

$$\Rightarrow I={\int }_{0}^{\pi }\frac{\mathrm{dx}}{1+2\mathrm{acosx}+{a}^{2}}...(ii)$$⇒I=∫0π​1+2acosx+a2dx​...(ii)

Adding $(i)$(i) and $(ii)$(ii),

$$\Rightarrow 2I={\int }_{0}^{\pi }\frac{\mathrm{dx}}{1-2\mathrm{acosx}+{a}^{2}}+{\int }_{0}^{\pi }\frac{\mathrm{dx}}{1+2\mathrm{acosx}+{a}^{2}}$$⇒2I=∫0π​1−2acosx+a2dx​+∫0π​1+2acosx+a2dx​

$$\Rightarrow 2I={\int }_{0}^{\pi }(\frac{1}{1-2\mathrm{acosx}+{a}^{2}}+\frac{1}{1+2\mathrm{acosx}+{a}^{2}})\mathrm{dx}$$⇒2I=∫0π​(1−2acosx+a21​+1+2acosx+a21​)dx

$$\Rightarrow 2I={\int }_{0}^{\pi }(\frac{1+2\mathrm{acosx}+{a}^{2}+1-2\mathrm{acosx}+{a}^{2}}{{(1+{a}^{2})}^{2}-{(2\mathrm{acosx})}^{2}})\mathrm{dx}$$⇒2I=∫0π​((1+a2)2−(2acosx)21+2acosx+a2+1−2acosx+a2​)dx

$$\Rightarrow 2I={\int }_{0}^{\pi }(\frac{2(1+{a}^{2})}{{(1+{a}^{2})}^{2}-{(2\mathrm{acosx})}^{2}})\mathrm{dx}$$⇒2I=∫0π​((1+a2)2−(2acosx)22(1+a2)​)dx

$$\Rightarrow 2I=2{\int }_{0}^{\frac{\pi }{2}}\frac{2(1+{a}^{2})}{{(1+{a}^{2})}^{2}-4{a}^{2}{\mathrm{cos}}^{2}x}\mathrm{dx}$$⇒2I=2∫02π​​(1+a2)2−4a2cos2x2(1+a2)​dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{2(1+{a}^{2})\cdot {\mathrm{sec}}^{2}x}{{(1+{a}^{2})}^{2}\cdot {\mathrm{sec}}^{2}x-4{a}^{2}}\mathrm{dx}$$⇒I=∫02π​​(1+a2)2⋅sec2x−4a22(1+a2)⋅sec2x​dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{2(1+{a}^{2})\cdot {\mathrm{sec}}^{2}x}{{(1+{a}^{2})}^{2}+{(1+{a}^{2})}^{2}{\mathrm{tan}}^{2}x-4{a}^{2}}\mathrm{dx}$$⇒I=∫02π​​(1+a2)2+(1+a2)2tan2x−4a22(1+a2)⋅sec2x​dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{2(1+{a}^{2})\cdot {\mathrm{sec}}^{2}x}{{(1+{a}^{2}-2a)}^{2}+{(1+{a}^{2})}^{2}{\mathrm{tan}}^{2}x}\mathrm{dx}$$⇒I=∫02π​​(1+a2−2a)2+(1+a2)2tan2x2(1+a2)⋅sec2x​dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{2\cdot (1+{a}^{2})\cdot {\mathrm{sec}}^{2}x}{{(1+{a}^{2})}^{2}\cdot {\mathrm{tan}}^{2}x+{(1-{a}^{2})}^{2}}\mathrm{dx}$$⇒I=∫02π​​(1+a2)2⋅tan2x+(1−a2)22⋅(1+a2)⋅sec2x​dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{\frac{2\cdot {\mathrm{sec}}^{2}x}{1+{a}^{2}}\cdot \mathrm{dx}}{{\mathrm{tan}}^{2}x+{(\frac{1-{a}^{2}}{1+{a}^{2}})}^{2}}$$⇒I=∫02π​​tan2x+(1+a21−a2​)21+a22⋅sec2x​⋅dx​

Let, $\mathrm{tanx}=t$tanx=t

$$\Rightarrow {\mathrm{sec}}^{2}\mathrm{xdx}=\mathrm{dt}$$⇒sec2xdx=dt

$$\Rightarrow I=\frac{2}{1+{a}^{2}}{\int }_{0}^{\infty }\frac{\mathrm{dt}}{{t}^{2}+{(\frac{1-{a}^{2}}{1+{a}^{2}})}^{2}}$$⇒I=1+a22​∫0∞​t2+(1+a21−a2​)2dt​

$$\Rightarrow I=\frac{2}{1+{a}^{2}}\times \frac{1}{(\frac{1-{a}^{2}}{1+{a}^{2}})}{[{\mathrm{tan}}^{-1}\frac{t}{(\frac{1-{a}^{2}}{1+{a}^{2}})}]}_{0}^{\infty }$$⇒I=1+a22​×(1+a21−a2​)1​[tan−1(1+a21−a2​)t​]0∞​

$$\Rightarrow I=\frac{2}{1-{a}^{2}}[{\mathrm{tan}}^{-1}(\infty )-{\mathrm{tan}}^{-1}(0)]$$⇒I=1−a22​[tan−1(∞)−tan−1(0)]

$$\Rightarrow I=\frac{2}{(1-{a}^{2})}[\frac{\pi }{2}-0]$$⇒I=(1−a2)2​[2π​−0]

$$\Rightarrow I=\frac{\pi }{1-{a}^{2}}$$⇒I=1−a2π​