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The integral intcos(logex)dx is equal t...

The integral `intcos(log_ex)dx ` is equal to (where C is a contant of integration)

A

`(x)/(2)[cos(log_(e)x)+sin(log_(e)x)]+C`

B

`x[cos(log_(e)x)+sin(log_(e)x)]+C`

C

`x[cos(log_(e)x)-sin(log_(e)x)]+C`

D

`(x)/(2)[cos(log_(e)x)-sin(log_(e)x)]+C`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `I=int cos(log_(e)x)dx`
`=x cos(log_(e)x)-intx(-sin(log_(e)x))(1)/(x)*dx" "["using integration by parts"]`
`= x cos(log_(e)x)+int sin(log_(e)x)dx`
`=x cos(log_(e)x)+ x sin(log_(e)x)-int x(cos(log_(e)x))(1)/(x)dx" "["again, using integration by parts"]`
`rArr I = x cos(log_(e)x)+x sin(log_(e)x)-I`
`rArr I=(x)/(2)[cos(log_(e)x)+sin(log_(e)x)]+C`.
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