A) \[\frac{\pi }{2ab}\]
B) \[\frac{\pi }{ab}\]
C) \[\frac{{{\pi }^{2}}}{2ab}\]
D) \[\frac{{{\pi }^{2}}}{ab}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi }{\frac{xdx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\] ?. (i) \[I=\int_{0}^{\pi }{\frac{(\pi -x)}{{{a}^{2}}{{\cos }^{2}}(\pi -x)+{{b}^{2}}{{\sin }^{2}}(\pi -x)}dx}\] \[\left[ Using\,\int_{0}^{a}{f(x)\,dx=}\int_{0}^{a}{f\,(a-x)\,dx} \right]\] or \[I=\int_{0}^{\pi }{\frac{(\pi -x)}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx}\] ... (ii) Adding (i) and (ii) \[2I=\int_{0}^{\pi }{\frac{x+\pi -x}{({{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x)}dx}\] \[=\pi \int_{0}^{\pi }{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\] \[=2\pi \int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\] \[\left[ \int_{0}^{2a}{f(x)\,dx=2}\int_{0}^{a}{f(x)}\,dx\,if\,(2a-x)=f(x) \right]\] = dividing numerator and denominator by \[{{\cos }^{2}}x\] \[2\pi \int_{0}^{\pi /2}{\frac{{{\sec }^{2}}x}{{{a}^{2}}+{{b}^{2}}{{\tan }^{2}}x}dx}\] Now put \[\tan x=t\] \[\therefore \] \[dx=\frac{dt}{{{\sec }^{2}}x}\] \[2\pi \int_{0}^{\infty }{\frac{dt}{{{a}^{2}}+{{b}^{2}}{{t}^{2}}}}\] \[=\frac{1}{{{b}^{2}}}\times 2\pi \int_{0}^{\infty }{\frac{dt}{\frac{{{a}^{2}}}{{{b}^{2}}}+{{t}^{2}}}}\] \[=\left[ \frac{2\pi }{b}\,.\,\frac{b}{a}{{\tan }^{-1}}\frac{bt}{a} \right]\] \[=\frac{2\pi }{ab}[{{\tan }^{-1}}\infty -{{\tan }^{-1}}0]\] \[=\frac{2\pi }{ab}\left( \frac{\pi }{2}-0 \right)\] \[=\frac{2\pi }{ab}\times \frac{\pi }{2}\] \[=\frac{2{{\pi }^{2}}}{2ab}=\frac{{{\pi }^{2}}}{ab}\]
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