int frac{1}{sqrt{8+2x-x^2}} dx = | vedclass

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

Question

(D) Let $I = \int \frac{dx}{\sqrt{8+2x-x^2}}$.First,complete the square for the quadratic expression inside the square root:$8+2x-x^2 = -(x^2-2x-8) =

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{x-1}{3}\right)+c$

Vedclass · Answer

(D) Let $I = \int \frac{dx}{\sqrt{8+2x-x^2}}$.First,complete the square for the quadratic expression inside the square root:$8+2x-x^2 = -(x^2-2x-8) = -(x^2-2x+1-9) = -( (x-1)^2 - 9 ) = 9 - (x-1)^2$.Now,substitute this back into the integral:$I = \int \frac{dx}{\sqrt{9-(x-1)^2}} = \int \frac{dx}{\sqrt{3^2-(x-1)^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right)+c$,we get:$I = \sin^{-1}\left(\frac{x-1}{3}\right)+c$.

$\int \frac{1}{\sqrt{8+2x-x^2}} dx =$

Similar Questions

$\int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}} } \;dx = $

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

If $\int(1-\cos x) \operatorname{cosec}^2 x \, dx = f(x) + c$,then $f(x)$ is equal to

$\int \frac{1 + \tan x \tan(x + a)}{\tan x \tan(x + a)} dx =$

If $\int(3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}) dt = f(t) \sin(\frac{1}{t}) + c$,then $f(2) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module