How do you evaluate the integral #int (sin^2x-cos^2x)/cosx# from #[-pi/4, pi/4]#?

1 Answer
Steve M
Nov 4, 2016

# int_(-pi/4)^(pi/4) (sin^2x-cos^2x)/cosx dx = ln( 3+2sqrt2 ) #

Explanation:

Using the fundamental trig Identity #sin^2A+cos^2A-=1# we have

# int_(-pi/4)^(pi/4) (sin^2x-cos^2x)/cosx dx = int_(-pi/4)^(pi/4) (1-cos^2x-cos^2x)/cosx dx #

# = int_(-pi/4)^(pi/4) (1-2cos^2x)/cosx dx #
# = int_(-pi/4)^(pi/4) 1/cosx-2cosx dx #
# = int_(-pi/4)^(pi/4) secx-2cosx dx #
# = [ln|secx+tanx| -2sinx]_(-pi/4)^(pi/4) #

# = (ln|sec(pi/4)+tan(pi/4)| -2sin(pi/4)) - (ln|sec(-pi/4)+tan(-pi/4)| -2sin(-pi/4)) #

# = (ln(sqrt2+1) -2 1/2 sqrt2) - (ln(sqrt2-1) -2 1/2 sqrt2) #
# = ln(sqrt2+1) -2 1/2 sqrt2 - ln(sqrt2-1) +2 1/2 sqrt2 #
# = ln(sqrt2+1) - ln(sqrt2-1) #
# = ln((sqrt2+1)/(sqrt2-1)) #
# = ln( (sqrt2+1)/(sqrt2-1) * (sqrt2+1)/(sqrt2+1) ) #
# = ln( (2+2sqrt2+1)/(2-1) ) #
# = ln( 3+2sqrt2 ) #

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