ExamHabit
Practise free

JEE Main 2019 · 9 Apr · Shift II

JEE Main 2019 Mathematics question on Indefinite Integration

The integral ∫sec⁡2/3xcosec⁡4/3x dx\int \sec^{2/3} x \cosec^{4/3} x \, dx is equal to (here CC is a constant of integration)
  1. A.
    3tan⁡−1/3x+C3\tan^{-1/3} x + C
  2. B.
    −3tan⁡−1/3x+C-3\tan^{-1/3} x + C
  3. C.
    −3cot⁡−1/3x+C-3\cot^{-1/3} x + C
  4. D.
    −34tan⁡−4/3x+C-\dfrac{3}{4}\tan^{-4/3} x + C
Show answer and solution

Correct option: B

Solution

Rewrite the integrand by dividing numerator and denominator: sec^(2/3)x · cosec^(4/3)x = (1/cos^(2/3)x)(1/sin^(4/3)x).

Divide top and bottom by cos2x to express in terms of tanx:

sec^(2/3)x · cosec^(4/3)x = (1/(sin^(4/3)x · cos^(2/3)x)) · (1/cos²x) · cos²x ... instead, write it directly:

sin^(4/3)x · cos^(2/3)x = tan^(4/3)x · cos²x, so the integrand = 1/(tan^(4/3)x · cos²x) = tan^(-4/3)x · sec²x.

So the integral becomes ∫ tan^(-4/3)x · sec²x dx. Let t = tanx, dt = sec²x dx:

∫ t^(-4/3) dt = t^(-1/3)/(-1/3) + C = -3t^(-1/3) + C = -3tan^(-1/3)x + C.

[JEE Main 2019 · 9 Apr · Shift II]

Practise Indefinite Integration the way it is asked

A timed set of real JEE Main Indefinite Integration questions, graded as the exam grades. No sign-up.

Start a Indefinite Integration test

More: all JEE Main Indefinite Integration questions · JEE Main 2019 · Mathematics chapter list