int_{-frac{pi}{2}}^{frac{pi}{2}} sin (x-[x]) | vedclass

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

Question

(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(x-[x]) \, dx$. Since $[x]$ is the greatest integer function,we split the integral at integer p

Vedclass · Accepted Answer

$2(1-\cos 1)$

Vedclass · Answer

(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(x-[x]) \, dx$. Since $[x]$ is the greatest integer function,we split the integral at integer points between $-\frac{\pi}{2} \approx -1.57$ and $\frac{\pi}{2} \approx 1.57$. The integer points are $-1, 0, 1$. $I = \int_{-\pi/2}^{-1} \sin(x-(-2)) \, dx + \int_{-1}^{0} \sin(x-(-1)) \, dx + \int_{0}^{1} \sin(x-0) \, dx + \int_{1}^{\pi/2} \sin(x-1) \, dx$. $I = \int_{-\pi/2}^{-1} \sin(x+2) \, dx + \int_{-1}^{0} \sin(x+1) \, dx + \int_{0}^{1} \sin(x) \, dx + \int_{1}^{\pi/2} \sin(x-1) \, dx$. Evaluating each integral: $1$. $[-\cos(x+2)]_{-\pi/2}^{-1} = -\cos(1) + \cos(2-\pi/2) = -\cos 1 + \sin 2$. $2$. $[-\cos(x+1)]_{-1}^{0} = -\cos 1 + \cos 0 = 1 - \cos 1$. $3$. $[-\cos x]_{0}^{1} = -\cos 1 + \cos 0 = 1 - \cos 1$. $4$. $[-\cos(x-1)]_{1}^{\pi/2} = -\cos(\pi/2-1) + \cos 0 = -\sin 1 + 1$. Summing these: $I = (\sin 2 - \cos 1) + (1 - \cos 1) + (1 - \cos 1) + (1 - \sin 1) = 3 - 3\cos 1 + \sin 2 - \sin 1$.

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

Similar Questions

$\int_{-2}^2 |x^2-x-2| dx =$

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$,then $b$ is equal to

If $\int_{n}^{n+1} f(x) dx = n^2 + n$ for all $n \in I$,then the value of $\int_{-3}^{3} f(x) dx$ is equal to

If $\int_{0}^{k} \frac{dx}{2 + 18x^2} = \frac{\pi}{24}$,then the value of $k$ is

The value of $\int_0^1 {x^2 e^x dx}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module