How do you integrate #sqrt(1- tan^2(x))# with respect to #x#?
2 Answers
Use WolframAlpha for complicated integrals, like this one.
Explanation:
I used WolframAlpha to do the integral.
Explanation:
#I=intsqrt(1-tan^2x)dx#
Let
#I=intsqrt(1-tan^2x)/sec^2x(sec^2xdx)#
#I=intsqrt(1-tan^2x)/(tan^2x+1)(sec^2xdx)#
#I=intsqrt(1-sin^2theta)/(1+sin^2theta)(costhetad theta)#
#I=intcos^2theta/(1+sin^2theta)d theta#
Write in terms of
#I=int(1/sec^2theta)/(1+tan^2theta/sec^2theta)d theta#
#I=int1/(sec^2theta+tan^2theta)d theta#
Let
#I=int1/(2tan^2theta+1)d theta#
Multiply by
#I=intsec^2theta/((tan^2theta+1)(2tan^2theta+1))d theta#
Let
#I=int1/((u^2+1)(2u^2+1))du#
Omitting the process of partial fraction decomposition, this becomes:
#I=2int1/(2u^2+1)du-int1/(u^2+1)du#
Both of which are forms of arctangent integrals:
#I=sqrt2intsqrt2/((sqrt2u)^2+1)du-int1/(u^2+1)du#
#I=sqrt2arctan(sqrt2u)-arctan(u)#
Recall
#I=sqrt2arctan(sqrt2tantheta)-arctan(tantheta)#
#I=sqrt2arctan(sqrt2tantheta)-theta#
And from
#I=sqrt2arctan((sqrt2tanx)/sqrt(1-tan^2x))-arcsin(tanx)+C#