Question

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

Answer 1

`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`.