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

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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

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

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)

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