How do you integrate #int thetasecthetatantheta# by parts?

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Answer 1 · 2017-02-04T08:44:28

#thetasectheta-ln|sectheta+tantheta|+C#

the parts formula

#intuv'd theta=uv-intvu'd theta#

#u=theta=>u'=1#

#v'=secthetatantheta=>v=sectheta#

#I=intuv'd theta=uv-intvu'd theta=thetasectheta-intsecthetad theta#

the problem now is to integrate #intsecthetad theta#

proceed as follows

#intsecthetad theta=int((sectheta)xx((sectheta+tantheta)/(sectheta+tantheta)))d theta#

#int((sec^2theta+secthetatantheta)/(sectheta+tantheta))d theta#

the numerator is the derivative of the denominator, so we have a log integral

#:.intsecthetad theta=ln|sectheta+tantheta|#

the original integral is now complete

#I=thetasectheta-intsecthetad theta#

#=>I=thetasectheta-ln|sectheta+tantheta|+C#