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

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$.

Let $f: R \rightarrow R$ be a continuous function such that $f(x)+f(x+1)=2$ for all $x \in R$. If $I_{1}=\int_{0}^{8} f(x) d x$ and $I_{2}=\int_{-1}^{3} f(x) d x$,then the value of $I_{1}+2 I_{2}$ is equal to:

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} d x$ is equal to

The value of the integral $\sum\limits_{k = 1}^n {\int_0^1 {f(k - 1 + x)\,dx} } $ is

The value of the integral $\int_{-\pi/4}^{\pi/4} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} \right) dx$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module