How do you integrate #sqrt(1- tan^2(x))# with respect to #x#?

2 Answers
Dec 10, 2016

Use WolframAlpha for complicated integrals, like this one.

Explanation:

I used WolframAlpha to do the integral.

#intsqrt(1 - tan^2(x))dx = #

# (cos(x) sqrt(1 - tan^2(x)) (sqrt(2) sin^(-1)(sqrt(2) sin(x)) - tan^(-1)((sin(x))/sqrt(cos(2 x)))))/sqrt(cos(2 x)) + C#

Mar 14, 2017

#sqrt2arctan((sqrt2tanx)/sqrt(1-tan^2x))-arcsin(tanx)+C#

Explanation:

#I=intsqrt(1-tan^2x)dx#

Let #sintheta=tanx# so #costhetad theta=sec^2xdx#.

#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 #tantheta# and #sectheta#:

#I=int(1/sec^2theta)/(1+tan^2theta/sec^2theta)d theta#

#I=int1/(sec^2theta+tan^2theta)d theta#

Let #sec^2theta=1+tan^2theta#:

#I=int1/(2tan^2theta+1)d theta#

Multiply by #sec^2theta/sec^2theta=sec^2theta/(tan^2theta+1)#:

#I=intsec^2theta/((tan^2theta+1)(2tan^2theta+1))d theta#

Let #u=tantheta# so #du=sec^2thetad theta#:

#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 #u=tantheta#:

#I=sqrt2arctan(sqrt2tantheta)-arctan(tantheta)#

#I=sqrt2arctan(sqrt2tantheta)-theta#

And from #sintheta=tanx# we note that #theta=arcsin(tanx)# and #tantheta=sintheta/sqrt(1-sin^2theta)#:

#I=sqrt2arctan((sqrt2tanx)/sqrt(1-tan^2x))-arcsin(tanx)+C#