Question

Evaluate the definite integrals.
${\int }_0^{\frac{\pi }{2}}\frac{co{s}^{2}x}{co{s}^{2}x+4si{n}^{2}x}dx.$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{co{s}^{2}x+4si{n}^{2}x}={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{4-3co{s}^{2}x}dx[\because si{n}^{2}x=1-co{s}^{2}x]=-\frac{1}{3}{\int }_0^{\frac{\pi }{2}}\frac{4-3co{s}^{2}x-4}{4-3co{s}^{2}x}dx=-\frac{1}{3}{\int }_0^{\frac{\pi }{2}}(1-\frac{4}{4-3co{s}^{2}x})dx=-\frac{1}{3}[x{]}_0^{\frac{\pi }{2}}+\frac{4}{3}{\int }_0^{\frac{\pi }{2}}\frac{se{c}^2}{4se{c}^{2}x-3}dx$
(Divide numerator and dinominator in second integral by $co{s}^{2}x$)
$=-\frac{\pi }{6}+\frac{4}{3}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{4(1+ta{n}^{2}x)-3}[\because 1+ta{n}^{2}x=se{c}^{2}x]Puttanx=t\Rightarrow se{c}^{2}dx=dtWhen,x=0\Rightarrow t=0andwhenx=\left(\frac{\pi }{2}\right)\Rightarrow t=\infty \therefore I=-\frac{\pi }{6}+\frac{4}{3}\times \frac{1}{4}{\int }_0^{\infty }\frac{dt}{t^2+{\left(\frac{1}{2}\right)}^2}=-\frac{\pi }{6}+\frac{1}{3}.2{\left[ta{n}^{-1}\frac{t}{\frac{1}{2}}\right]}_0^{\infty }(\because \int \frac{dx}{a^2+x^2}=\frac{1}{a}ta{n}^{-1}\frac{x}{a})=-\frac{\pi }{6}+\frac{2}{3}[ta{n}^{-1}2t{]}_0^{\infty }=-\frac{\pi }{6}+\frac{2}{3}[ta{n}^{-1}\infty -ta{n}^{-1}0]=-\frac{\pi }{6}+\frac{2}{3}(\frac{\pi }{2}-0)=-\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{6}$