What is int1/(a+cos(x))*dx where a is a constant?
1 Answer
Explanation:
We will rewrite
cos(2alpha)=2cos^2(alpha)-1
Let
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
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
I=2int1/(a+1+(a-1)u^2)du
Hopefully we can see an inverse tangent integral in the making.
Let
This implies that
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
I=2/(sqrt(a-1)sqrt(a+1))intd theta=2/sqrt(a^2-1)theta+C
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