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inte^(x)(sinx+cosx)dx=?...

`inte^(x)(sinx+cosx)dx=?`

A

`e^(x)sinx+C`

B

`e^(x)cosx+C`

C

`e^(x)tanx+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`I=inte^(x){f(x)+f'(x)}dx," where "f(x)=secx`
`=e^(x)secx+C`.
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