JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
The integral
∫_(π/6)^(π/3) tan ³x.sin ²3x( 2sec ²x.sin ²3x + 3 tan x.sin 6x )dx
is equal to:
- (a) - 1/9
- (b) - 1/18
- (c) 7/18
- (d) 9/2
Answer and solution
Answer (b) - 1/18
Solution
Maths sir · 4 steps
The bracket is almost the derivative of tan⁴ x sin⁴ 3x, so the integral is just two endpoint values.
- Expand the integrand. Use sin 6x = 2 sin 3x cos 3x. The integrand becomes 2 tan³ x sec² x sin⁴ 3x + 6 tan⁴ x sin³ 3x cos 3x.
- Spot the derivative. By the product rule, d/dx(tan⁴ x sin⁴ 3x) = 4 tan³ x sec² x sin⁴ 3x + 12 tan⁴ x sin³ 3x cos 3x. That is exactly twice our integrand.
I = 1/2[tan⁴ x sin⁴ 3x]_(π/6)^(π/3)
- Put in limits. At x = π/3: sin π = 0, so the value is 0. At x = π/6: tan⁴ π/6 = 1/9 and sin π/2 = 1, so the value is 1/9.
- So. I = 1/2(0 - 1/9) = -1/18, which is option (b).
The idea to keep. When an integrand has powers of two functions times their derivatives, guess fᵐ gⁿ and differentiate it. Match the coefficients, and the integral becomes plain endpoint values.
Answer from the NTA final answer key. Solution by PrepWiser.