intlimits_0^pi {{e^{{{cos }^4}x}}} cdot cos^5 | vedclass

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

Question

(D) Let $I = \int\limits_0^\pi e^{\cos^4 x} \cdot \cos^5(2n + 1)x \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx$,we have: $I

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $I = \int\limits_0^\pi e^{\cos^4 x} \cdot \cos^5(2n + 1)x \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx$,we have: $I = \int_0^\pi e^{\cos^4(\pi - x)} \cdot \cos^5(2n + 1)(\pi - x) \,dx$. Since $\cos(\pi - x) = -\cos x$,then $\cos^4(\pi - x) = \cos^4 x$. Also,$\cos(2n + 1)(\pi - x) = \cos((2n + 1)\pi - (2n + 1)x) = \cos((2n + 1)\pi) \cdot \cos(2n + 1)x + \sin((2n + 1)\pi) \cdot \sin(2n + 1)x$. Since $(2n + 1)$ is an odd integer,$\cos((2n + 1)\pi) = -1$ and $\sin((2n + 1)\pi) = 0$. Thus,$\cos(2n + 1)(\pi - x) = -\cos(2n + 1)x$. Therefore,$\cos^5(2n + 1)(\pi - x) = (-\cos(2n + 1)x)^5 = -\cos^5(2n + 1)x$. Substituting this back into the integral: $I = \int_0^\pi e^{\cos^4 x} \cdot (-\cos^5(2n + 1)x) \,dx = -I$. $2I = 0 \implies I = 0$.

$\int\limits_0^\pi {{e^{{{\cos }^4}x}}} \cdot \cos^5(2n + 1)x \,dx, (n \in I)$ is equal to

Similar Questions

The value of $\int_{0}^{\pi / 2} \log (\operatorname{cosec} x) d x$ is

$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = $ . . . . . . .

The value of the integral $I = \int_{0}^{1} x(1 - x)^n dx$ is

Which of the following are true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module