What is the the value of int sqrt(tanx/(sinxcosx)) dx?

1 Answer
Dec 17, 2016

= ln|secx + tanx| + C

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 secx+ tanx.

=int(secx xx (secx + tanx)/(secx + tanx))dx

=int(sec^2x + secxtanx)/(secx + tanx)dx

Let u = secx + tanx. Then (du)/dx = sec^2x + secxtanx and du = sec^2x + secxtanx dx. Substitute:

=int(du)/u

=ln|u| + C

= ln|secx + tanx| + C

Hopefully this helps!