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 2x \cos^3(x^2) \sin(x^2) \, dx = $

If $\int \sin ^5 x \, dx = \frac{-\cos ^5 x}{5} + a \cos ^3 x + b \cos x + c$,then $a + b =$

$f(x) = \int \frac{dx}{\sin^6 x}$ is a polynomial of degree

$\begin{aligned} & \int \frac{dx}{(2 \sin x+\sec x)^4}=A(1+\tan x)^{-5} \\ & +B(1+\tan x)^{-6}+C(1+\tan x)^{-7}+k, \text{ then } \\ & A+B+C= \end{aligned}$

$\int \frac{(\sin^{-1} x)^{\frac{3}{2}}}{\sqrt{1-x^2}} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module