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)

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

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

(C) $\frac{1}{2} \cos \sqrt{(2 x+1)^{2}+5}+c$

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

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Indefinite Integrals

The integral

\int \frac{( 2 x - 1 ) c o s ( 2 x - 1 ) ^{2} + 5}{4 x ^{2} - 4 x + 6} d x

is equal to : (where c is a constant of integration)

Answer: (A)

Since

4 x^{2} - 4 x + 6 = (2 x - 1)^{2} + 5

, the radical in the denominator is the same radical that appears inside the cosine. Put

u = (2 x - 1)^{2} + 5, u^{2} = (2 x - 1)^{2} + 5

Differentiating,

2 u d u = 4 (2 x - 1) d x

, so

(2 x - 1) d x = \frac{u d u}{2}

and the integral becomes

\int \frac{cos u}{u} \cdot \frac{u d u}{2} = \frac{1}{2} \int cos u d u = \frac{1}{2} sin u + c

Restoring

u

, the integral equals

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

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option