Evaluate int_{alpha}^{beta} sqrt{frac{x-alpha | vedclass

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

Question

(C) Let $I = \int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$.Substitute $x = \alpha \cos^2 t + \beta \sin^2 t$.Then $dx = 2(\beta - \alpha)

Vedclass · Accepted Answer

$\frac{\pi}{2}(\beta - \alpha)$

Vedclass · Answer

(C) Let $I = \int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$.Substitute $x = \alpha \cos^2 t + \beta \sin^2 t$.Then $dx = 2(\beta - \alpha) \sin t \cos t dt$.When $x = \alpha$,$t = 0$. When $x = \beta$,$t = \frac{\pi}{2}$.$x - \alpha = (\beta - \alpha) \sin^2 t$ and $\beta - x = (\beta - \alpha) \cos^2 t$.So,$\sqrt{\frac{x-\alpha}{\beta-x}} = \sqrt{\frac{(\beta-\alpha)\sin^2 t}{(\beta-\alpha)\cos^2 t}} = 	an t$.$I = \int_{0}^{\pi/2} 	an t \cdot 2(\beta - \alpha) \sin t \cos t dt = 2(\beta - \alpha) \int_{0}^{\pi/2} \sin^2 t dt$.Using the identity $\sin^2 t = \frac{1 - \cos 2t}{2}$,we get:$I = 2(\beta - \alpha) \int_{0}^{\pi/2} \frac{1 - \cos 2t}{2} dt = (\beta - \alpha) [t - \frac{\sin 2t}{2}]_{0}^{\pi/2}$.$I = (\beta - \alpha) [\frac{\pi}{2} - 0] = \frac{\pi}{2}(\beta - \alpha)$.

Evaluate $\int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$

Similar Questions

$\int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \log_e(\sin x + \cos x) \, dx$ is equal to

The value of $b > 3$ for which $12 \int \limits_{3}^{b} \frac{1}{(x^{2}-1)(x^{2}-4)} dx = \log _{e}(\frac{49}{40})$ is equal to

$\int_{0}^{\pi/4} \sqrt{1+\sin 2x} dx = \rule{1cm}{0.15mm}$

Evaluate the definite integral $\int_{0}^{1} \left(x e^{x} + \sin \frac{\pi x}{4}\right) dx$.

Evaluate the definite integral $\int_{-2}^{2.24} [x] \, dx$,where $[x]$ denotes the greatest integer function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module