What is the the value of #int sqrt(tanx/(sinxcosx)) dx#?
1 Answer
Dec 17, 2016
Explanation:
Start by simplifying the expression within the integral.
#=intsqrt((sinx/cosx)/(sinxcosx))dx#
#=intsqrt((sinx)/(cosxsinxcosx))dx#
#=intsqrt(1/cos^2x)dx#
#=int(1/cosx)dx#
#=int(secx)dx#
This is a tricky integral to do. Multiply everything by
#=int(secx xx (secx + tanx)/(secx + tanx))dx#
#=int(sec^2x + secxtanx)/(secx + tanx)dx#
Let
#=int(du)/u #
#=ln|u| + C#
#= ln|secx + tanx| + C#
Hopefully this helps!