To find the value of int frac{dx}{xsqrt{2ax - | vedclass

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

Question

(D) To evaluate the integral $I = \int \frac{dx}{x\sqrt{2ax - x^2}}$,we use the substitution $x = \frac{2a}{t}$.Then,$dx = -\frac{2a}{t^2} dt$.Substit

Vedclass · Accepted Answer

$x = \frac{2a}{t}$

Vedclass · Answer

(D) To evaluate the integral $I = \int \frac{dx}{x\sqrt{2ax - x^2}}$,we use the substitution $x = \frac{2a}{t}$.Then,$dx = -\frac{2a}{t^2} dt$.Substituting these into the integral:$I = \int \frac{-\frac{2a}{t^2} dt}{\frac{2a}{t} \sqrt{2a(\frac{2a}{t}) - (\frac{2a}{t})^2}}$$I = \int \frac{-\frac{2a}{t^2} dt}{\frac{2a}{t} \sqrt{\frac{4a^2}{t} - \frac{4a^2}{t^2}}}$$I = \int \frac{-\frac{2a}{t^2} dt}{\frac{2a}{t} \cdot \frac{2a}{t} \sqrt{t - 1}}$$I = -\int \frac{dt}{t \sqrt{t - 1}}$.This substitution simplifies the integral into a standard form. Note that the original option $(d)$ in the prompt was $x = 2a \sin^2 t$,which is also a valid substitution,but $x = \frac{2a}{t}$ is the standard substitution for this specific form.

To find the value of $\int \frac{dx}{x\sqrt{2ax - x^2}}$,the suitable substitution is

Similar Questions

$\int {{e^{ - x}}{{\csc }^2}(2{e^{ - x}} + 5)} \,dx = $

$\int \frac{1}{x(\log x)^2} \, dx = $

Find the value of $k$ if $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.

Integrate the function: $\frac{\sin x}{\sin (x-a)}$

$\int (1+e^{-x})^{-1} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module