JEE Main 2019MathematicsIndefinite IntegrationMediumMCQ

JEE Main 2019 — Indefinite Integration Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

For, x2≠nπ+1, n∈N (the set of natural numbers), the integral ∫x2sinx2-1-sin2x2-12sinx2-1+sin2x2-1dx, is equal to
(where c is a constant of integration).

Choose an option

Show full solutionCorrect option: D
Correct answer
Dlogesec2x2-12+c

Step-by-step explanation

⇒I=∫x2sinx2-1-sin2x2-12sinx2-1+sin2x2-1dx

Let, x2-1=θ⇒xdx=12dθ

⇒I=12∫2sinθ-sin2θ2sinθ+sin2θdθ

=12∫2sinθ-2sinθcosθ2sinθ+2sinθcosθdθ

=12∫1-cosθ1+cosθdθ

=12∫tanθ2dθ

=12logesecθ212+c, where c is the constant of integration.

=logesec2x2-12+c

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About this question

This is a previous-year question from JEE Main 2019, covering the Indefinite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.