tan32xsec2x = tan22xtan2xsec2x= (sec22x -1)tan2xsec2x= sec22x.tan2xsec2x-tan2xsec2x⇒ ⇒ Now, Let sec2x = t⇒ 2sec2xtan2x dx = dtThus, ⇒ ⇒

Q. 153.8( 86 Votes )

Find the integral

Answer :

tan³2xsec2x = tan²2xtan2xsec2x

= (sec²2x -1)tan2xsec2x

= sec²2x.tan2xsec2x-tan2xsec2x

⇒

Now, Let sec2x = t

⇒ 2sec2xtan2x dx = dt

Thus,

⇒

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