Question-107617 Tinku Tara June 4, 2023 None 0 Comments FacebookTweetPin Question Number 107617 by mathdave last updated on 11/Aug/20 Answered by Ar Brandon last updated on 11/Aug/20 I=∫sec3xtan3xdx=∫sec2x(sec2x−1)secxtanxdx=∫(sec4x−sec2x)d(secx)=[sec5x5−sec3x3]+C Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 0-1-n-x-arcsin-x-dx-Next Next post: If-x-2-2-2-3-2-1-3-then-prove-that-x-3-6x-2-6x-2-0- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![I=∫sec^3 xtan^3 xdx =∫sec^2 x(sec^2 x−1)secxtanxdx =∫(sec^4 x−sec^2 x)d(secx) =[((sec^5 x)/5)−((sec^3 x)/3)]+C](https://www.tinkutara.com/question/Q107622.png)