The value of the integral int_{1/pi }^{2/pi } | vedclass

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

Question

(D) Let $t = \frac{1}{x}$.Then,$dt = -\frac{1}{x^2} \,dx$,which implies $-\,dt = \frac{1}{x^2} \,dx$.When $x = \frac{1}{\pi}$,$t = \pi$.When $x = \fra

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Let $t = \frac{1}{x}$.Then,$dt = -\frac{1}{x^2} \,dx$,which implies $-\,dt = \frac{1}{x^2} \,dx$.When $x = \frac{1}{\pi}$,$t = \pi$.When $x = \frac{2}{\pi}$,$t = \frac{\pi}{2}$.Substituting these into the integral:$\int_{1/\pi}^{2/\pi} \frac{\sin(1/x)}{x^2} \,dx = \int_{\pi}^{\pi/2} \sin(t) \cdot (-dt)$$= \int_{\pi/2}^{\pi} \sin(t) \,dt$$= [-\cos(t)]_{\pi/2}^{\pi}$$= -(\cos(\pi) - \cos(\pi/2))$$= -(-1 - 0) = 1$.

The value of the integral $\int_{1/\pi }^{2/\pi } \frac{\sin(1/x)}{x^2} \,dx$ is:

Similar Questions

$\int_0^{\frac{\pi}{4}} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} \, dx =$

$\int\limits_0^{{{\left( {\frac{\pi }{2}} \right)}^{\frac{1}{3}}}} {\,{x^5}\cdot\sin {x^3}\,dx} $ $=$

Let $\frac{d}{dx}F(x) = \frac{e^{\sin x}}{x}$ for $x > 0$. If $\int_{1}^{4} \frac{3}{x} e^{\sin(x^3)} dx = F(k) - F(1)$,then one of the possible values of $k$ is:

$\int_{1}^{e} \frac{dx}{x(1+\log x)^{2}} =$

If $\frac{d[f(x)]}{dx} = g(x)$ for $a \le x \le b$,then $\int_a^b f(x)g(x) dx$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module