int frac{1}{sqrt{2x-x^2}} dx = . . . . . . | vedclass

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

Question

(B) To solve the integral $I = \int \frac{1}{\sqrt{2x-x^2}} dx$,we first complete the square for the expression inside the square root: $2x - x^2 = -(

Vedclass · Accepted Answer

$\sin^{-1}(x-1)$

Vedclass · Answer

(B) To solve the integral $I = \int \frac{1}{\sqrt{2x-x^2}} dx$,we first complete the square for the expression inside the square root: $2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$. Substituting this back into the integral,we get: $I = \int \frac{1}{\sqrt{1 - (x-1)^2}} dx$. Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}(\frac{u}{a}) + C$,where $u = x-1$ and $a = 1$: $I = \sin^{-1}(\frac{x-1}{1}) + C = \sin^{-1}(x-1) + C$. Therefore,the correct option is $B$.

$\int \frac{1}{\sqrt{2x-x^2}} dx = $ . . . . . . $+ C$.

Similar Questions

Integrate the function: $\sqrt{4-x^{2}}$

$\int {\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \dots} \right) dx} = $

$\int \frac{dx}{(\sin x)(\cos x)}$ is equal to

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} \, dx$

If $\frac{3 \pi}{4} < x < \frac{7 \pi}{4}$, then $\int \left(2^x - \sqrt{1 + \sin 2x} + \frac{1}{x^2} - \frac{1}{x}\right) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module