For any integer n, the integral int_0^pi {{e^ | vedclass

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

Question

(B) Let $I = \int_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}(2n + 1)x\,dx}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^\

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $I = \int_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}(2n + 1)x\,dx}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^\pi {{e^{{{\sin }^2}(\pi-x)}}{{\cos }^3}(2n + 1)(\pi-x)\,dx}$.Since $\sin(\pi-x) = \sin x$,we have $\sin^2(\pi-x) = \sin^2 x$.Also,$\cos((2n+1)(\pi-x)) = \cos((2n+1)\pi - (2n+1)x) = -\cos((2n+1)x)$ because $(2n+1)$ is an odd integer.Therefore,$\cos^3((2n+1)(\pi-x)) = (-\cos((2n+1)x))^3 = -\cos^3((2n+1)x)$.Substituting these into the integral:$I = \int_0^\pi {{e^{{{\sin }^2}x}} \cdot [-\cos^3(2n + 1)x]\,dx} = -I$.Thus,$2I = 0$,which implies $I = 0$.

For any integer $n,$ the integral $\int_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}(2n + 1)x\,dx} = $

Similar Questions

Let $T > 0$ be a fixed number. Suppose $f$ is a continuous function such that for all $x \in \mathbb{R}, f(x + T) = f(x)$. If $I = \int_{0}^{T} f(x) dx$,then the value of $\int_{3}^{3 + 3T} f(2x) dx$ is

Let $A = \int\limits_0^1 \frac{e^t}{1 + t} \, dt$. Then $\int\limits_{a - 1}^a \frac{e^{-t}}{t - a - 1} \, dt$ has the value:

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(x^5-x^3 \cos x+\sin^3 x-3) \, dx = $ . . . . . .

The value of the integral $\int_{-2}^{2} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \, dx$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is

The value of $\int_1^3 \sqrt{3 + x^3} \,dx$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module