JEE Main 2019MathematicsIndefinite IntegrationEasyMCQ

JEE Main 2019 — Indefinite Integration Question with Solution

JEE Main 2019 (10 Jan Shift 1)

Question

Let, n≥2 be a natural number and 0<θ<π2. Then ∫sinnθ-sinθ1ncosθsinn+1θdθ, is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
Cnn2-11-1sinn-1θn+1n+c

Step-by-step explanation

∫sinnθ-sinθ1ncosθsinn+1θdθ

Put, sinθ=t⇒cosθdθ=dt

 =∫tn-t1ndttn+1

=∫t1-1tn-11ntn+1dt

= ∫1-1tn-11ntndt

Put 1-1tn-1=z

⇒n-1tndt=dz

⇒I=1n-1∫z1ndz

Using ∫xndx=xn+1n+1+c

⇒I=z1n+11n+1n-1+c

⇒I=n1-t1-n1n+1n2-1+c, where c is the constant of integration.
⇒I=nn2-11-1sinn-1θn+1n+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.