\hspace{-16}$Calculate value of \\\\\\ $\mathbf{(1)::I_{1}=\int_{0}^{\pi}\frac{\sin x}{7+4\cos 2x+\cos^2 2x}dx}$\\\\\\ $\mathbf{(2)::I_{2}=\int_{0}^{\pi}\frac{x.\sin x}{7+4\cos 2x+\cos^2 2x}dx}$\\\\
\hspace{-16}$Ans\; :: $I_{1}= \frac{\pi\sqrt{3}+3\ln(3)}{24}$\\\\ $I_{2}= \frac{\pi}{2}.\left(\frac{\pi\sqrt{3}+3\ln(3)}{24}\right)$\\\\
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1 Answers
1. \int_{0}^{\pi}{\frac{\sin xdx}{7+4\cos2x+\cos^22x}}
=\frac{1}{4}\int_{-1}^{1}{\frac{dt}{t^4+t^{2}+1}}=\frac{1}{2}\int_{0}^{1}{\frac{dt}{t^4+t^{2}+1}}
\text{Now, } \frac{1}{t^4+t^2+1}=\frac{At+B}{t^2+t+1}+\frac{Ct+D}{t^2-t+1}
\text{Which gives } A=\frac{1}{2},B=\frac{1}{2},C=-\frac{1}{2},D=\frac{1}{2}
Now, the integral reduces to well-known forms and evalutes to \boxed{\frac{\pi\sqrt{3}+3\ln3}{24}} .
2. \: I=\int_{0}^{\pi}{\frac{x\sin xdx}{7+4\cos2x+\cos^22x}}
\text{Also, }I=\int_{0}^{\pi}{\frac{\left( \pi-x\right)\sin xdx}{7+4\cos2x+\cos^22x}}
\therefore \: 2I= \pi\int_{0}^{\pi}{\frac{\sin xdx}{7+4\cos2x+\cos^22x}}
\Rightarrow I=\frac{\pi}{2}\int_{0}^{\pi}{\frac{\sin xdx}{7+4\cos2x+\cos^22x}}=\boxed{\frac{\pi}{2}\left(\frac{\pi\sqrt{3}+3\ln3}{24} \right)}