The parametric form of a curve is x = frac{t^ | vedclass

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

Question

(A) Given $x = \frac{t^3}{t^2 - 1}$ and $y = \frac{t}{t^2 - 1}$.First,calculate $x - 3y$:$x - 3y = \frac{t^3}{t^2 - 1} - 3 \left( \frac{t}{t^2 - 1} \r

Vedclass · Accepted Answer

$\frac{1}{2} \log(t^2 - 1) + C$

Vedclass · Answer

(A) Given $x = \frac{t^3}{t^2 - 1}$ and $y = \frac{t}{t^2 - 1}$.First,calculate $x - 3y$:$x - 3y = \frac{t^3}{t^2 - 1} - 3 \left( \frac{t}{t^2 - 1} \right) = \frac{t^3 - 3t}{t^2 - 1}$.Next,find $dx$ by differentiating $x$ with respect to $t$:$dx = \frac{d}{dt} \left( \frac{t^3}{t^2 - 1} \right) dt = \frac{(t^2 - 1)(3t^2) - t^3(2t)}{(t^2 - 1)^2} dt = \frac{3t^4 - 3t^2 - 2t^4}{(t^2 - 1)^2} dt = \frac{t^4 - 3t^2}{(t^2 - 1)^2} dt$.Now,substitute these into the integral:$\int \frac{dx}{x - 3y} = \int \frac{\frac{t^4 - 3t^2}{(t^2 - 1)^2} dt}{\frac{t^3 - 3t}{t^2 - 1}} = \int \frac{t^2(t^2 - 3)}{(t^2 - 1)^2} \cdot \frac{t^2 - 1}{t(t^2 - 3)} dt = \int \frac{t}{t^2 - 1} dt$.Let $u = t^2 - 1$,then $du = 2t dt$,so $t dt = \frac{1}{2} du$.$\int \frac{1}{2u} du = \frac{1}{2} \log|u| + C = \frac{1}{2} \log(t^2 - 1) + C$.

The parametric form of a curve is $x = \frac{t^3}{t^2 - 1}$,$y = \frac{t}{t^2 - 1}$,then $\int \frac{dx}{x - 3y} =$

Similar Questions

Integrate the function $\frac{x}{e^{x^{2}}}$.

$\int \frac{x}{\sqrt{x+4}} \, dx = $ . . . . . . $+ C, x > -4$.

$\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = $

For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,if $y(x) = \int \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x} \, dx$ and $\lim_{x \rightarrow (\frac{\pi}{2})^-} y(x) = 0$,then $y\left(\frac{\pi}{4}\right)$ is equal to:

Evaluate the integral: $\int \sec^2 \theta (\sec \theta + \tan \theta)^2 d\theta$

Vedclass Test Series

Exam Paper Generator

Online Exam Module