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int[1+tanx.tan(x+alpha)]dx is equal to...

`int[1+tanx.tan(x+alpha)]dx` is equal to

A

`cosalpha.ln|(sinx)/((sin(x+alpha)+C))|`

B

`tanalpha.ln|(sinx)/(sin(x+alpha)+C)|`

C

`cotalpha.ln|(sec(x+alpha))/(secx)|+C`

D

`cotalpha.ln|(cos(x+alpha))/(cosx)|+C`

Text Solution

Verified by Experts

The correct Answer is:
C
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