Question

the value of $\sum _{m=1}^{10}{I}_{2m+1}$ is equal to

• A. $0$
• B. $5\pi$
• C. $10\pi$
• D. None of these

Answer 1

The correct option is C $10\pi$

Given $I_n={\int }_{-\pi }^{\pi }\frac{sinnx}{(1+{\pi }^x)sinx}dx...\left(i\right)$

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

$I_n={\int }_{-\pi }^{\pi }\frac{{\pi }^{x}sinnx}{(1+{\pi }^x)sinx}dx..\left(ii\right)$

On adding Eqs.(i) and (ii), we have

$2I_n={\int }_{-\pi }^{\pi }\frac{sinnx}{sinx}dx=2{\int }_0^{\pi }\frac{sinnx}{sinx}dx[\because f(x)=\frac{sinnx}{sinx}$ is an even function]

$\Rightarrow I_n={\int }_0^{\pi }\frac{sinnx}{sinx}dx$

$\Rightarrow I_1=\pi$ and $I_2=0$

From Eq.(iii) $I_1=I_3=I_5=...=\pi$

and $I_2=I_4=I_6=...=0$

$\Rightarrow \sum _{m=1}^{10}{I}_{2m+1}=10\pi$