Mathematics - Definite Integration Question with Solution | TestHub

(a) $$\begin{array}{rl}& I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{\left(1+{\pi }^x\right)\sin x}dx\\ & I_n={\int }_{-\pi }^{\pi }\frac{{\pi }^x\sin nx}{\left(1+{\pi }^x\right)\sin x}dx\\ & \left[\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx\right]\\ & 2I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{\sin x}dx\end{array}$$ $$\begin{array}{rl}& \Rightarrow 2I_n=2{\int }_0^{\pi }\frac{\sin nx}{\sin x}dx\\ & \Rightarrow I_n={\int }_0^{\pi }\frac{\sin nx}{\sin x}dx\\ & \text{ Now, }{I}_{n+2}-I_n\\ & ={\int }_0^{\pi }\frac{\sin (n+2)x-\sin nx}{\sin x}dx\\ & ={\int }_0^{\pi }\frac{2\cos (n+1)x\sin x}{\sin x}dx\\ & =2{\left[\frac{\sin (n+1)x}{(n+1)}\right]}_0^{\pi }=0\\ & \Rightarrow I_{n+2}=I_n\\ & \text{ (b) As }{I}_3=I_5=\dots =I_{21}\\ & \therefore \sum _{m=1}^{10}{I}_{2m+1}=10I_3=10{\int }_0^{\pi }\frac{\sin 3x}{\sin x}dx\\ & =10{\int }_0^{\pi }\left(3-4{\sin }^{2}x\right)dx\\ & =10[3x-2x+\sin 2x{]}_0^{\pi }=10\pi \\ & \text{ (c) As }{I}_2=I_4=\dots =I_{20}\\ & \therefore \sum _{m=1}^{10}{I}_{2m}=10{\int }_0^{\pi }\frac{\sin 2x}{\sin x}dx\\ & =20[\sin x{]}_0^{\pi }=0\\ & \end{array}$$