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  • Evaluate the definite integrals in Exercises 25 to 33. 27.

Evaluate the definite integrals in Exercises 25 to 33.

    Q27.    \int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}

Answers (1)

best_answer

Lets first simplify the function.

\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4(1-\cos^2 x)}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{4-3\cos^2 x}

\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x-4\:\: }{4-3\cos^2 x }dx=\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x\:\: }{4-3\cos^2 x }dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx

\\=\frac{-1}{3}\int_0^\frac{\pi}{2}1dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx \\ \\ \\=\frac{-1}{3} \left [ x \right ]_0^{\frac{\pi}{2}}-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4\:\: }{4-3\cos^2 x }dx

 

As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4sec^2x-3 }dx

AS we can write square of sec in term of tan,

 

=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4(1+tan^2x)-3 }dx

=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx

Now let's calculate the integral of the second function, (we already have calculated the first function)

=-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx

let

 \\t=2tanx, \\dt=2sec^2xdx

here we are changing the variable so we have to calculate the limits of the new variable

when x = 0, t = 2tanx = 2tan(0)=0

whenx=\pi/2,t=2tan{\pi/2}=\infty

our function in terms of t is 

 

=-\frac{2}{3}\int_0^\infty\frac{1 }{1+t^2 }dt

=\left [ tan^{-1} t\right ]_0^\infty=\left [ tan^{-1} \infty-tan^{-1} 0\right ]

=\frac{\pi}{2}

Hence our total solution of the function is 

\\=-\frac{\pi}{6}+\frac{2}{3}*\frac{\pi}{2}\\=\frac{\pi}{6}

 

Posted by

Pankaj Sanodiya

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