int_{-pi}^{frac{pi}{2}} sin x cdot sin^2(cos | vedclass

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

Question

(C) Let $I = \int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx$. Substitute $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$

Vedclass · Accepted Answer

$\frac{\sin 2-2}{4}$

Vedclass · Answer

(C) Let $I = \int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx$. Substitute $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$. When $x = -\pi$,$t = \cos(-\pi) = -1$. When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$. Thus,$I = \int_{-1}^{0} \sin^2(t) (-dt) = \int_{0}^{-1} \sin^2(t) \, dt$. Using the identity $\sin^2(t) = \frac{1 - \cos(2t)}{2}$,we get: $I = \int_{0}^{-1} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \left[ t - \frac{\sin(2t)}{2} \right]_{0}^{-1}$. $I = \frac{1}{2} \left[ (-1 - \frac{\sin(-2)}{2}) - (0 - 0) \right]$. Since $\sin(-2) = -\sin(2)$,we have: $I = \frac{1}{2} \left[ -1 + \frac{\sin(2)}{2} \right] = \frac{\sin(2) - 2}{4}$.

$\int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx =$

Similar Questions

Prove that $\int_{0}^{\frac{\pi}{4}} 2 \tan^{3} x \, dx = 1 - \log 2$.

$\int_{5}^{10} \frac{1}{(x-1)(x-2)} d x$ is equal to

If $f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \\ 4x^3 - 1, & x > 1 \end{cases}$, then $\int_{0}^{2} f(x) dx$ is

Let $f: R \rightarrow R$ be defined as $f(x)=e^{-x} \sin x$. If $F :[0,1] \rightarrow R$ is a differentiable function such that $F(x)=\int_{0}^{x} f(t) dt$,then the value of $\int_{0}^{1}(F'(x)+f(x)) e^{x} dx$ lies in the interval

$\int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2}+4 \sin \frac{x}{2}} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module