int_{frac{-3pi}{2}}^{frac{-pi}{2}} left[ (x+p | vedclass

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

Question

(C) Let $I = \int_{-\frac{3\pi}{2}}^{-\frac{\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx$. Substitute $t = x + \pi$,then $dt = dx$. When $x =

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(C) Let $I = \int_{-\frac{3\pi}{2}}^{-\frac{\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx$. Substitute $t = x + \pi$,then $dt = dx$. When $x = -\frac{3\pi}{2}$,$t = -\frac{\pi}{2}$. When $x = -\frac{\pi}{2}$,$t = \frac{\pi}{2}$. Since $\cos(x+3\pi) = \cos(x+\pi) = -\cos x$,we have $\cos^2(x+3\pi) = \cos^2 x$. Thus,$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (t^3 + \cos^2 t) dt$. Since $t^3$ is an odd function and $\cos^2 t$ is an even function,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t^3 dt = 0$. Therefore,$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 t dt = 2 \int_{0}^{\frac{\pi}{2}} \cos^2 t dt$. Using the identity $\cos^2 t = \frac{1 + \cos 2t}{2}$,we get $I = 2 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2t}{2} dt = \int_{0}^{\frac{\pi}{2}} (1 + \cos 2t) dt$. Evaluating the integral: $[t + \frac{\sin 2t}{2}]_{0}^{\frac{\pi}{2}} = (\frac{\pi}{2} + 0) - (0 + 0) = \frac{\pi}{2}$.

$\int_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx = $

Similar Questions

The value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \frac{\pi-x}{\pi+x}\right) \cos x \, dx$ is equal to

If ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta ,} $ then for any positive integer $n,$ the value of $n({I_{n - 1}} + {I_{n + 1}})$ is

$\int_0^\pi \frac{x \, dx}{1 + \sin x}$ is equal to

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$,$x \in R - \{-1\}$,$n \in N$,$n > 2$. If $f^n(x) = (f \circ f \circ f \dots \text{upto } n \text{ times})(x)$,then $\lim_{n \to \infty} \int_0^1 x^{n-2} (f^n(x)) dx$ is equal to $...............$.

$\int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx = a \ln 2 + b$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module