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

Explanation

`int_0^(pi//6)cos^(-3)2thetasin2thetad theta`

`I=int_0^(pi//6)cos^(-3)2thetasin2thetad theta`

`I=int_0^(pi//6)(tan2thetasec^2(2theta)2d theta)/2`

`I=(1/2)int_0^(pi//6)tan2thetasec^2(2theta)d theta`

As we know `int tanxsec^2xdx=(tan^2x)/2+C`

Therefore, `(1/2)int_0^(pi//6)tan2 theta sec^2(2theta)d theta=(1/2)[(tan^2(2theta))/2]_0^(pi/6)`

`I=(1/4)[tan^2(pi/3)−0]`

`I=(3/4)`