Mathematics › Indefinite Integration › Integration by Substitution
Let,
I=∫1+tan2x1-tan2xdx
⇒I=∫sec2x1-tan2xdx
Put tanx=t, we get
sec2x dx=dt
I=∫dt1-t2
⇒I=12log 1+t1-t+c
⇒I=12log 1+tanx1-tanx+c.
Asked in: AP EAMCET 2020 (23 Sep Shift 1)
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