If int frac{1}{sqrt{9-16 x^2}} d x=alpha sin | vedclass

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

Question

(A) We are given the integral $\int \frac{1}{\sqrt{9-16 x^2}} d x$. Rewrite the denominator as $\sqrt{3^2-(4 x)^2}$. Using the standard formula $\int

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) We are given the integral $\int \frac{1}{\sqrt{9-16 x^2}} d x$. Rewrite the denominator as $\sqrt{3^2-(4 x)^2}$. Using the standard formula $\int \frac{1}{\sqrt{a^2-u^2}} d u = \sin^{-1}(\frac{u}{a}) + c$,where $u = 4x$ and $du = 4 dx$,we get: $\int \frac{1}{\sqrt{3^2-(4 x)^2}} d x = \frac{1}{4} \int \frac{1}{\sqrt{3^2-(4 x)^2}} d(4x) = \frac{1}{4} \sin^{-1}(\frac{4x}{3}) + c$. Comparing this with $\alpha \sin^{-1}(\beta x) + c$,we identify $\alpha = \frac{1}{4}$ and $\beta = \frac{4}{3}$. Now,calculate $\alpha + \frac{1}{\beta} = \frac{1}{4} + \frac{1}{4/3} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$.

If $\int \frac{1}{\sqrt{9-16 x^2}} d x=\alpha \sin ^{-1}(\beta x)+c$,then $\alpha+\frac{1}{\beta}=$

Similar Questions

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

$\int \tan^4 x \, dx = $

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

If $\int \frac{x}{x \tan x+1} \, dx = \log f(x) + k$,then $f\left(\frac{\pi}{4}\right) =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module