What is #int1/(a+cos(x))*dx# where #a# is a constant?

We will rewrite #cos(x)# using some identities. Starting with the cosine double angle formula:

#cos(2alpha)=2cos^2(alpha)-1#

Let #alpha=x/2# to show that

#cos(x)=2cos^2(x/2)-1#

Rewriting:

#cos(x)=2/sec^2(x/2)-1=(2-sec^2(x/2))/sec^2(x/2)#

The denominator of the integrand is then:

#a+cos(x)=a+(2-sec^2(x/2))/sec^2(x/2)=(2+(a-1)sec^2(x/2))/sec^2(x/2)#

In the numerator, let #sec^2(x/2)=tan^2(x/2)+1#:

#a+cos(x)=(2+(a-1)(tan^2(x/2)+1))/sec^2(x/2)#

#color(white)(a+cos(x))=(a+1+(a-1)tan^2(x/2))/sec^2(x/2)#

So:

#I=int1/(a+cos(x))dx=intsec^2(x/2)/(a+1+(a-1)tan^2(x/2))dx#

Let #u=tan(x/2)#. This implies that #du=1/2sec^2(x/2)dx#:

#I=2int1/(a+1+(a-1)u^2)du#

Hopefully we can see an inverse tangent integral in the making.

Let #(a-1)u^2=(a+1)tan^2(theta)#.

This implies that #sqrt(a+1)tan(theta)=sqrt(a-1)(u)#, so #sqrt(a+1)sec^2(theta)d theta=sqrt(a-1)du#.

Then:

#I=2/sqrt(a-1)int1/(a+1+(a+1)tan^2(theta))sqrt(a+1)sec^2(theta)d theta#

#color(white)(I)=(2sqrt(a+1))/(sqrt(a-1)(a+1))intsec^2(theta)/(1+tan^2(theta))d theta#

Since #1+tan^2(theta)=sec^2(theta)#:

#I=2/(sqrt(a-1)sqrt(a+1))intd theta=2/sqrt(a^2-1)theta+C#

#sqrt(a+1)tantheta=sqrt(a-1)(u)# implies that #theta=tan^-1(usqrt((a-1)/(a+1)))#:

#I=2/sqrt(a^2-1)tan^-1(usqrt((a-1)/(a+1)))+C#

#color(white)(I)=2/sqrt(a^2-1)tan^-1(tan(x/2)sqrt((a-1)/(a+1)))+C#

Which is only valid when #(a+1)(a-1)gt0#, or for all values of #a# except #-1lt=alt=1#.