Question

${\int }_0^{\pi }\frac{xdx}{4{\cos }^{2}x+9{\sin }^{2}x}=$

• A. $\frac{{\pi }^2}{12}$
• B. $\frac{{\pi }^2}{4}$
• C. $\frac{{\pi }^2}{6}$
• D. $\frac{{\pi }^2}{3}$

Answer 1

The correct option is A $\frac{{\pi }^2}{12}$

To find the Integral of

${\int }_0^{\Pi }\frac{xdx}{4co{s}^{2}x+9si{n}^{2}x}$

Let ,

$I={\int }_0^{\Pi }\frac{xdx}{4co{s}^{2}x+9si{n}^{2}x}$ --------> Equation 1

As we know that ,

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^{\Pi }\frac{(\Pi -x)dx}{4co{s}^2(\Pi -x)+9si{n}^2(\Pi -x)}$

$I={\int }_0^{\Pi }\frac{(\Pi -x)dx}{4co{s}^{2}x+9si{n}^{2}x}$ ------> Equation 2

On Adding Equation 1 and 2 we get

$2I={\int }_0^{\Pi }\frac{(\Pi -x+x)dx}{4co{s}^{2}x+9si{n}^{2}x}$

$I=\frac{\Pi }{2}{\int }_0^{\Pi }\frac{dx}{4co{s}^{2}x+9si{n}^{2}x}$

$I=\frac{\Pi }{18}{\int }_0^{\Pi }\frac{dx}{\frac{4}{9}co{s}^{2}x+si{n}^{2}x}$

$I=\frac{\Pi }{18}{\int }_0^{\Pi }\frac{se{c}^{2}xdx}{ta{n}^{2}x+\frac{4}{9}}$

$I=\frac{2\Pi }{18}{\int }_0^{\Pi /2}\frac{se{c}^{2}xdx}{(tanx{)}^2+{\left(\frac{2}{3}\right)}^2}$

$I=\frac{\Pi }{9}{\int }_0^{\Pi /2}\frac{se{c}^{2}xdx}{(tanx{)}^2+{\left(\frac{2}{3}\right)}^2}$

Let $tanx=t$

So that $se{c}^{2}xdx=dt$

When $x=0,t=0$

When $x=\frac{\Pi }{2}$ , $t=\infty$

Above Integral Becomes ,

$I=\frac{\Pi }{9}{\int }_0^{\infty }\frac{dt}{t^2+{\left(\frac{2}{3}\right)}^2}$

$I=\frac{\Pi }{6}{\left[ta{n}^{-1}\left\{\frac{t}{\frac{2}{3}}\right\}\right]}_{t=0}^{\infty }$

$I=\frac{\Pi }{6}{\left[ta{n}^{-1}\left(\frac{3t}{2}\right)\right]}_{t=0}^{\infty }$

$I=\frac{\Pi }{6}\times [ta{n}^{-1}\infty -ta{n}^{-1}(0\left)\right]$

$I=\frac{\Pi }{6}\times [\frac{\Pi }{2}-0]$

$I=\frac{\Pi }{6}\times \frac{\Pi }{2}$

$I=\frac{{\Pi }^2}{12}$