Question

If $f\left(x\right)=\frac{{\sin }^2\mathrm{\pi x}}{1+{\mathrm{\pi }}^x}$. Then, $\int \left[f\left(x\right)+f(-x)\right]dx$ is equal to

• A. $0$
• B. $x+C$
• C. $\left(\frac{x}{2}\right)-\left(\frac{\mathrm{cos\pi }x}{2\mathrm{\pi }}\right)+C$
• D. $\left[\frac{1}{\left(1+\mathrm{\pi }\right)}\right]\left[\left(\frac{{\cos }^2\mathrm{\pi x}}{2\mathrm{\pi }}\right)\right]+C$
• E. $\frac{x}{2}-\frac{\sin 2\mathrm{\pi x}}{4\mathrm{\pi }}+C$

Answer 1

The correct option is E

$\frac{x}{2}-\frac{\sin 2\mathrm{\pi x}}{4\mathrm{\pi }}+C$

Explanation for the correct option:

Find the value of $\int \left[f\left(x\right)+f(-x)\right]dx$:

Given,

$f\left(x\right)=\frac{{\sin }^2\mathrm{\pi x}}{1+{\mathrm{\pi }}^x}...\left(1\right)$

$\begin{array}{rcl}f(-x)& =& \frac{{\sin }^2\left(-\pi x\right)}{1+{\mathrm{\pi }}^{-x}}\\ & =& \frac{{\mathrm{\pi }}^x{\sin }^2\left(-\pi x\right)}{1+{\mathrm{\pi }}^x}...\left(2\right)\end{array}$

Substitute equations $\left(1\right)$ and $\left(2\right)$ in $\int \left[f\left(x\right)+f(-x)\right]dx$.

$\begin{array}{rcl}\int \left[f\left(x\right)+f(-x)\right]dx& =& \int \left[\left(\frac{{\sin }^2\pi x}{1+{\mathrm{\pi }}^x}\right)+\left(\frac{{\mathrm{\pi }}^x{\sin }^2\mathrm{\pi }x}{1+{\mathrm{\pi }}^x}\right)\right]dx\\ & =& \int \left[\left(\frac{{\sin }^2\pi x}{1+{\mathrm{\pi }}^x}\right)\left(1+{\mathrm{\pi }}^x\right)\right]dx\\ & =& \int {\sin }^2\pi xdx\\ & =& \frac{1}{2}\int \left(1-\cos 2\mathrm{\pi }x\right)dx\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\pi }\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{\pi }\mathbf{x}\mathbf{]}\\ & =& \frac{1}{2}\left(x-\frac{\sin 2\mathrm{\pi }x}{2\mathrm{\pi }}\right)+C\\ & =& \frac{x}{2}-\frac{\sin 2\mathrm{\pi }x}{4\mathrm{\pi }}+C\end{array}$

Hence, the correct option is E.