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

1 Answer
Apr 21, 2017

int1/(a+cos(x))dx=2/sqrt(a^2-1)tan^-1(tan(x/2)sqrt((a-1)/(a+1)))+C

Explanation:

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.