Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The indefinite integral I=sec2xtanxsecx+tanxdxsec5x+sec2xtan3x-sec3xtan2x-tan5x simplifies to 13lnfx+c, where fπ4=22+1 and c is the constant of integration. If the value of fπ3 is a+b, then the value of b-3a is equal to
MathematicsIndefinite IntegrationJEE Main
Solution:
2106 Upvotes Verified Answer
The correct answer is: 3

I=133sec3xtanx+3sec2xtan2xdxsec2x-tan2xsec3x+tan3x
Let sec3x+tan3x=t
Then, I=13dtt
=13lnt+c
=13lnsec3x+tan3x+c
fx=sec3x+tan3x
fπ3=8+27

a=8, b=27
b-3a=3

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.