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Grade:

                        

tanx.sec2x.dx=

8 years ago

Answers : (1)

Rinkoo Gupta
askIITians Faculty
80 Points
							
integral tanx.sec2x dx
=integral sinx/(cosx.cos2x) dx
=integral sinx/cosx(2cos^2x-1) dx
put cosx =t
on diff. w.r. to x
-sinx dx=dt
=>integral (-dt)/t(2t^2-1)
Using Partial fraction we get
integral (1/t -1/root2(t.root2.-1) -1/root2.(troot2+1))
=logt-1/2loh(troot2-1)-1/2 .log(troot2+1)
=log cosx-1/2log(2cos^2x-1)
=logcosx -1/2 logcos2x
=log (cosx/ root of cos2x)

Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
6 years ago
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