JEE Main 2016MathematicsDefinite IntegrationMediumMCQ

JEE Main 2016 — Definite Integration Question with Solution

JEE Main 2016 (09 Apr Online)

Question

If 2∫01tan-1xdx=∫01cot-11-x+x2dx, then ∫01tan-11-x+x2dx is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
Bln2

Step-by-step explanation

From the given equation

2∫01tan-1⁡xdx= ∫01π2-tan-1⁡1-x+x2dx

⇒2∫01tan-1⁡xdx= ∫01π2 dx- ∫01tan-1⁡1-x+x2dx
 ⇒∫01tan-1⁡1-x+x2dx=π2-2 ∫01tan-1⁡xdx    …i         
Let I1= ∫01tan-1⁡xdx

=tan-1⁡xx01-∫0111+x2xdx

=π4- ∫01x1+x2dx

=π4-12 ∫012x1+x2dx

=π4-12 ln1+x201

=π4-12ln⁡2

By equation i,

∫01tan-1⁡1-x+x2dx= π2-2π4-12ln2=ln⁡2 

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About this question

This is a previous-year question from JEE Main 2016, covering the Definite 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.