JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

If ∫ (cos xdx)/sin³x(1+sin⁶x)^(2/3)=f(x)(1+sin⁶x)^(1/λ)+c
where c is a constant of integration, then λ f(π/3) is equal to

  1. (a) 9/8
  2. (b) 2
  3. (c) -2
  4. (d) -9/8
Answer and solution

Answer (c) -2

Solution

Maths sir · 4 steps

Put t = sin x, then pull t⁶ out of the bracket so the derivative of the inside appears.

  1. Substitute sin x. Let t = sin x, so dt = cos x dx. The integral becomes ∫ dt/t³(1+t⁶)^(2/3).
  2. Factor out the power. Write (1+t⁶)^(2/3) = t⁴(1+t⁻⁶)^(2/3). The integrand becomes t⁻⁷/(1+t⁻⁶)^(2/3), and t⁻⁷ is exactly what the derivative of 1+t⁻⁶ needs.

    u = 1 + t⁻⁶, du = -6t⁻⁷ dt ⇒ -1/6∫ u^(-2/3) du = -1/2u^(1/3) + c

  3. Back to x. u^(1/3) = ((1+t⁶)/t⁶)^(1/3) = (1+t⁶)^(1/3)/t². So the integral is -1/(2 sin² x)(1+sin⁶ x)^(1/3) + c, giving f(x) = -1/(2 sin² x) and λ = 3.
  4. So. sin² π/3 = 3/4, so f(π/3) = -1/(2 × 3/4) = -2/3. Then λ f(π/3) = 3 × (-2/3) = -2, option (c).

The idea to keep. When the bracket has (1 + tⁿ) under a fractional power, take tⁿ outside. The leftover power of t often matches the derivative of 1 + t⁻ⁿ.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Indefinite Integration

Asked about 60 times in JEE Main since 2020. This question is from 2020.