int_{1/4}^{1/2} frac{dx}{sqrt{x - x^2}} = | vedclass

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

Question

(D) Let $I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}}$.We can rewrite the expression inside the square root by completing the square:$x - x^2 = -(x^

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(D) Let $I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}}$.We can rewrite the expression inside the square root by completing the square:$x - x^2 = -(x^2 - x) = -((x - 1/2)^2 - 1/4) = (1/2)^2 - (x - 1/2)^2$.Thus,$I = \int_{1/4}^{1/2} \frac{dx}{\sqrt{(1/2)^2 - (x - 1/2)^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,we get:$I = [\sin^{-1}(\frac{x - 1/2}{1/2})]_{1/4}^{1/2} = [\sin^{-1}(2x - 1)]_{1/4}^{1/2}$.Evaluating at the limits:$I = \sin^{-1}(2(1/2) - 1) - \sin^{-1}(2(1/4) - 1) = \sin^{-1}(0) - \sin^{-1}(-1/2)$.$I = 0 - (-\pi/6) = \pi/6$.

$\int_{1/4}^{1/2} \frac{dx}{\sqrt{x - x^2}} = $

Similar Questions

$\int_{0}^{\infty} \frac{x \, dx}{(1 + x)(1 + x^2)} = $

Evaluate the definite integral $\int_{-1}^{1}(x+1) d x$.

Let $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} dx$ $(x \geq 0)$. Then $f(3) - f(1)$ is equal to

If $f(x) = \begin{cases} e^{\cos x} \sin x, & \text{for } |x| \leq 2 \\ 2, & \text{otherwise} \end{cases}$,then $\int_{-2}^{3} f(x) dx$ is equal to

The function $L(x) = \int_1^x \frac{dt}{t}$ satisfies the equation

Vedclass Test Series

Exam Paper Generator

Online Exam Module