int frac{dx}{sqrt{4x-9x^2}} = rule{1cm}{0.15m | vedclass

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

Question

(B) To evaluate the integral $I = \int \frac{dx}{\sqrt{4x-9x^2}}$, we first factor out $9$ from the square root:$I = \int \frac{dx}{\sqrt{9(\frac{4}{9

Vedclass · Accepted Answer

$\frac{1}{3} \sin^{-1} \left( \frac{9x-2}{2} \right)$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{dx}{\sqrt{4x-9x^2}}$, we first factor out $9$ from the square root:$I = \int \frac{dx}{\sqrt{9(\frac{4}{9}x - x^2)}} = \frac{1}{3} \int \frac{dx}{\sqrt{\frac{4}{9}x - x^2}}$.Next, we complete the square for the quadratic expression inside the square root:$\frac{4}{9}x - x^2 = -(x^2 - \frac{4}{9}x) = -((x - \frac{2}{9})^2 - (\frac{2}{9})^2) = (\frac{2}{9})^2 - (x - \frac{2}{9})^2$.Substituting this back into the integral:$I = \frac{1}{3} \int \frac{dx}{\sqrt{(\frac{2}{9})^2 - (x - \frac{2}{9})^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C$, we get:$I = \frac{1}{3} \sin^{-1}\left( \frac{x - 2/9}{2/9} \right) + C = \frac{1}{3} \sin^{-1}\left( \frac{9x - 2}{2} \right) + C$.

$\int \frac{dx}{\sqrt{4x-9x^2}} = \rule{1cm}{0.15mm} + C$

Similar Questions

$\int \frac{dx}{\sqrt{x} + \sqrt{x - 2}} = $

$\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} = $

$\sqrt{2} \int \frac{\sin x \, dx}{\sin \left( x - \frac{\pi}{4} \right)} = $

$\int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \,d x$ is equal to

$\int \sqrt{\frac{1+x}{1-x}} \, dx = $ (where $C$ is a constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module