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

2 Answers
Apr 11, 2018

# = 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)#

Apr 11, 2018

#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)#