`intx.sinx.sec^(3)xdx=`

A

(a) `1/2(sec^(2)x-tanx)+c`

B

(b) `1/2(xsec^(2)x-tanx)+c`

C

(c) `1/2(xsec^(2)x+tanx)+c`

D

(d) `1/2(sec^(2)x+tanx)+c`

Verified by Experts

The correct Answer is:

B

Similar Questions

Explore conceptually related problems

intsin(11x).sin^(9)xdx="_______"+c .

int xcos^(2)xdx=

intx^(2)secx^(3)dx=

intx^(2)logxdx=

intsinxcos^(4)xdx=

intsec^6xdx=

intsin^(-1)xdx

intxe^(x)dx=?

intxsin2xdx=?