`intsin^3thetad theta`

intcot^4thetad theta

int_0^pisin^8thetad theta

From the first fundamental principle of integral calculs, we get f'(x)xsinx Using proper substitution, evaluate int_(0)^((pi)/(6))(1-cos3theta)sin3thetad theta

int_0^(pi/2)sin^10thetad theta

int_0^(pi/2)sin^11thetad theta

int_(-pi)^(pi)sin^(2)thetad theta is equal to-

int_(0)^((pi)/(4))tan^(3)thetad theta

Evaluate the following integrals: int_0^(pi//6)cos^(-3)2thetasin2thetad theta

intsec^(3)thetad""theta

The integral I=intsin (2theta)[(1+cos^(2)theta)/(2sin^(2)theta)]d theta simplifies to (where, c is the integration constant)