int_0^{1/2} frac{x sin^{-1} x}{sqrt{1 - x^2}} | vedclass

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

Question

(B) Let $t = \sin^{-1} x$. Then $dt = \frac{1}{\sqrt{1 - x^2}} dx$. When $x = 0$,$t = 0$. When $x = 1/2$,$t = \sin^{-1}(1/2) = \pi/6$. Also,$x = \sin

Vedclass · Accepted Answer

$\frac{1}{2} - \frac{\sqrt{3} \pi}{12}$

Vedclass · Answer

(B) Let $t = \sin^{-1} x$. Then $dt = \frac{1}{\sqrt{1 - x^2}} dx$. When $x = 0$,$t = 0$. When $x = 1/2$,$t = \sin^{-1}(1/2) = \pi/6$. Also,$x = \sin t$. Substituting these into the integral: $\int_0^{\pi/6} t \sin t \, dt$. Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = t$,$dv = \sin t \, dt$. Then $du = dt$,$v = -\cos t$. $\int t \sin t \, dt = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t$. Evaluating the definite integral: $[-t \cos t + \sin t]_0^{\pi/6} = (-\frac{\pi}{6} \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})) - (0 + 0)$. $= -\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1}{2} - \frac{\sqrt{3} \pi}{12}$.

$\int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = $

Similar Questions

Let $f:(0, \infty) \rightarrow \mathbb{R}$ and $F(x)=\int_0^x t f(t) d t$. If $F(x^2)=x^4+x^5$,then $\sum_{r=1}^{12} f(r^2)$ is equal to :

$\int_0^{\pi /4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ equals

If $f(x) = A\sin \left( \frac{\pi x}{2} \right) + B$,$f'\left( \frac{1}{2} \right) = \sqrt{2}$ and $\int_0^1 f(x) \, dx = \frac{2A}{\pi}$,then the constants $A$ and $B$ are respectively:

$\int_{\pi /6}^{\pi /4} \text{cosec} \, 2x \, dx = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module