If f(x) = frac{sin^2 pi x}{1+pi^x},then int ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We are given $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$. \ We need to evaluate $I = \int (f(x) + f(-x)) \, dx$. \ First,calculate $f(-x)$: \ $f(-x)

Vedclass · Accepted Answer

$\frac{1}{2} x - \frac{\sin 2 \pi x}{4 \pi} + c$,(where $c$ is a constant of integration)

Vedclass · Answer

(B) We are given $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$. \ We need to evaluate $I = \int (f(x) + f(-x)) \, dx$. \ First,calculate $f(-x)$: \ $f(-x) = \frac{\sin^2(-\pi x)}{1+\pi^{-x}} = \frac{\sin^2(\pi x)}{1+\frac{1}{\pi^x}} = \frac{\pi^x \sin^2 \pi x}{\pi^x + 1}$. \ Now,add $f(x)$ and $f(-x)$: \ $f(x) + f(-x) = \frac{\sin^2 \pi x}{1+\pi^x} + \frac{\pi^x \sin^2 \pi x}{1+\pi^x} = \frac{\sin^2 \pi x (1+\pi^x)}{1+\pi^x} = \sin^2 \pi x$. \ Now,integrate: \ $I = \int \sin^2 \pi x \, dx = \int \frac{1 - \cos 2 \pi x}{2} \, dx$. \ $I = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos 2 \pi x \, dx$. \ $I = \frac{x}{2} - \frac{1}{2} \cdot \frac{\sin 2 \pi x}{2 \pi} + c = \frac{x}{2} - \frac{\sin 2 \pi x}{4 \pi} + c$.

If $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$,then $\int (f(x) + f(-x)) \, dx$ is equal to

Similar Questions

$\int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x=$

$y=\int \cos \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x$ is an equation of a family of

Let $x > 0$ be a fixed real number. Then,the integral $\int \limits_0^{\infty} e^{-t}|x-t| d t$ is equal to

If $f(x) = \int x^2 \cos^2 x (2x \tan^2 x - 2x - 6 \tan x) dx$ and $f(0) = \pi$,then $f(x) =$

$\int \sqrt{1 + \csc x} \, dx$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module