JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
The integral ∫ ((2x-1)cos √((2x-1)²+5))/√(4x²-4x+6)dx is equal to (where c is a constant of integration)
- (a) 1/2 sin √((2x-1)²+5)+c
- (b) 1/2 cos √((2x+1)²+5)+c
- (c) 1/2 cos √((2x-1)²+5)+c
- (d) 1/2 sin √((2x+1)²+5)+c
Answer and solution
Answer (a) 1/2 sin √((2x-1)²+5)+c
Solution
Maths sir · 3 steps
The denominator is the same square root sitting inside the cosine, so put the whole root equal to t.
- Complete the square. 4x² - 4x + 6 = (4x² - 4x + 1) + 5 = (2x-1)² + 5. So the denominator is √((2x-1)² + 5), the same root as inside the cosine.
- Substitute. Let t = √((2x-1)² + 5), so t² = (2x-1)² + 5. Differentiating both sides gives 2t dt = 4(2x-1) dx, that is, ((2x-1) dx)/t = dt/2.
∫ ((2x-1)cos t)/t dx = 1/2∫ cos t dt = 1/2 sin t + c
- So. Putting t back, the integral is 1/2 sin√((2x-1)² + 5) + c, which is option (a).
The idea to keep. When the same messy expression appears inside a function and also in the denominator, substitute the whole expression. Complete the square first so you can see the match.
Answer from the NTA final answer key. Solution by PrepWiser.