Integration of 1/(a+btanx)

[bookworm189]

How do you integrate

1 / (a + b tan x)

 

[Deleted member 4993]

How do you integrate

1 / (a + b tan x)

One of the ways:

1/[a + b*tan(x)] = cos(x)/[a*cos(x) + b * sin(x)]

let

a/√(a² + b²) = sin(Φ) and

b/√(a² + b²) = cos(Φ)

then

1/[a + b*tan(x)] = 1/√(a² + b²) * {cos(x) / [sin(Φ)*cos(x) + cos(Φ) * sin(x)]} = 1/√(a² + b²) * {cos(x) / sin(Φ + x)}

Now continue.....

 

[galactus]

Here is another way to tackle it. This method can be handy on various integrals of the form

\(\displaystyle \int\frac{a\cdot \cos(x)+b\sin(x)}{c\cdot \cos(x)+e\sin(x))}dx\)

by letting \(\displaystyle \text{Numer}=A[\text{Denom}]+B\frac{d}{dx}[\text{Denom}]\).

\(\displaystyle \int\frac{1}{a+b\tan(x)}dx\)

\(\displaystyle =\int \frac{1}{a+b\frac{\sin(x)}{\cos(x)}}dx\)

\(\displaystyle =\int\frac{\cos(x)}{a\cdot \cos(x)+b\sin(x)}dx\)

Let \(\displaystyle \cos(x)=A(a\cdot \cos(x)+b\cdot \sin(x))+B\frac{d}{dx}(a\cdot \cos(x)+b\cdot \sin(x))\)

\(\displaystyle \cos(x)=A(a\cdot\cos(x)+b\sin(x))+B(-a\cdot \sin(x)+b\cos(x))\)

Equate coefficients of cos(x):

\(\displaystyle 1=Aa+Bb\rightarrow Aa+Bb-1=0\)

Equate coefficients of sin(x):

\(\displaystyle Ab-Ba=0\)

Solve the system for A and B in terms of a and b:

\(\displaystyle A=\frac{a}{a^{2}+b^{2}}, \;\ B=\frac{b}{a^{2}+b^{2}}\)

So, we get: \(\displaystyle \int\frac{A(a\cdot \cos(x)+b\sin(x))+B(-a\cdot sin(x)+b\cos(x))}{a\cdot \cos(x)+b\sin(x)}dx\)

\(\displaystyle =\int\left[A+B\cdot \frac{-a\cdot \sin(x)+b\cos(x)}{a\cdot \cos(x)+b\sin(x)}\right]dx\)

\(\displaystyle =Ax+B\log(a\cdot \cos(x)+b\sin(x))=\frac{ax}{a^{2}+b^{2}}+\frac{b}{a^{2}+b^{2}}\log(a\cdot \cos(x)+b\sin(x))\)

 

[Deleted member 4993]

Bookworm,

Also try wolframalpha.com - they use (basically) Galactus's method.

If this problem was assigned by a teacher - s/he does not like you very much!!!

 

[bookworm189]

Thank you so much!