The integral int frac{(2 x-1) cos sqrt{(2 x-1 | vedclass

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

Question

(A) Let $I = \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$. Note that $4x^2 - 4x + 6 = (2x-1)^2 + 5$. So,$I = \int \frac{(2

Vedclass · Accepted Answer

$\frac{1}{2} \sin \sqrt{(2 x-1)^{2}+5}+c$

Vedclass · Answer

(A) Let $I = \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$. Note that $4x^2 - 4x + 6 = (2x-1)^2 + 5$. So,$I = \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{(2 x-1)^{2}+5}} d x$. Let $u = \sqrt{(2 x-1)^{2}+5}$. Then $u^2 = (2x-1)^2 + 5$. Differentiating both sides with respect to $x$,we get $2u \frac{du}{dx} = 2(2x-1) \cdot 2 = 4(2x-1)$. Thus,$(2x-1) dx = \frac{u}{2} du$. Substituting these into the integral: $I = \int \frac{\cos u}{u} \cdot \frac{u}{2} du = \frac{1}{2} \int \cos u du$. $I = \frac{1}{2} \sin u + c$. Substituting back $u = \sqrt{(2 x-1)^{2}+5}$,we get $I = \frac{1}{2} \sin \sqrt{(2 x-1)^{2}+5} + c$.

The integral $\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$ is equal to (where $c$ is a constant of integration)

Similar Questions

$\int \frac{x^{3} \sin \left(\tan ^{-1}\left(x^{4}\right)\right)}{1+x^{8}} d x$ is equal to

$\int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} = $

$\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$ is equal to,where $c$ is an arbitrary constant.

To evaluate $\int \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \, dx$,the most suitable substitution is

$\int \frac{\cos x + x \sin x}{x(x - \cos x)} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module