JEE PYQ: Indefinite Integration Question 2
Question 2 - 2021 (18 Mar Shift 1)
The integral $\int \frac{(2x-1)\cos\sqrt{(2x-1)^2+5}}{\sqrt{4x^2 - 4x + 6}},dx$ is equal to (where $c$ is a constant of integration):
(1) $\frac{1}{2}\sin\sqrt{(2x-1)^2 + 5} + c$
(2) $\frac{1}{2}\cos\sqrt{(2x+1)^2 + 5} + c$
(3) $\frac{1}{2}\cos\sqrt{(2x-1)^2 + 5} + c$
(4) $\frac{1}{2}\sin\sqrt{(2x+1)^2 + 5} + c$
Show Answer
Answer: (1)
Solution
Let $(2x-1)^2 + 5 = t^2$, so $2(2x-1)\cdot 2,dx = 2t,dt$. The integral becomes $\frac{1}{2}\int \cos t,dt = \frac{1}{2}\sin t + c = \frac{1}{2}\sin\sqrt{(2x-1)^2+5} + c$.
BETA
