Question

How do you find the definite integral for: `int_(pi/4)^(pi/3)sin(2p) / (1+cos^2(p)) dp`?

Answer 1

`= ln (6/5)`

Explanation:

`int_(pi/4)^(pi/3)sin(2p) / (1+cos^2(p)) dp`

`= int_(pi/4)^(pi/3)(2 sin(p)cos (p)) / (1+cos^2(p)) dp`

Spotting the pattern, the numerator looks suspiciously like the derivative of the denominator:

`= - int_(pi/4)^(pi/3) d (ln(1+cos^2(p)))`

Reversing interval using minus sign

`= [ ln(1+cos^2(p)) ]_(pi/3)^(pi/4)`

`= ln(1+1/2) - ln(1+1/4) = ln (6/5)`

Answer 2

`int_(pi/4)^(pi/3)sin(2p)/(1+cos^2 2p)dp=1/2arctan(1/2)`

Explanation:

We want to find `int_(pi/4)^(pi/3)sin(2p)/(1+cos^2 2p)dp`.

We start by making a substitution

Let `u=cos2p` and `du=-2sin2pdu`.

Then `u(pi/3)=-1/2` and `u(pi/4)=0`

Substituting all of this gives an integral

`int_(pi/4)^(pi/3)sin(2p)/(1+cos^2 2p)dp=-1/2int_(0)^(-1/2)1/(1+u^2)du=1/2int_(-1/2)^(0)1/(1+u^2)du=1/2[arctanu]_(-1/2)^(0)=1/2(arctan(0)-arctan1/2))=1/2arctan(1/2)`