Evaluate: `int_(-pi)^(3pi)log(s e c theta-t a ntheta)d theta`

Explanation

Let `I=int_(-pi)^(3pi)log(sec theta-tan theta)d theta`…………..1
Using the property `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`, we get
`I=int_(-pi)^(3pi)log[sec (2pi-theta)-tan(2pi-theta)]d theta`
`int_(-pi)^(3pi) log[sec theta+tan theta]d theta`.................2
Adding 1 and 2 we get
`2I=int_(-pi)^(3pi)log[(sec theta-tan theta )(sec theta+tan theta)]d theta`
`=int_(-pi)^(3pi) log(1)d theta=int_(-pi)^(3pi)0, d theta=0` or `I=0`