Question

Integrate (1)/(sinx +4cosx)dx?

Answer 1

`1/(sinx+cosx)dx = -1/sqrt17ln(csc(x+arctan4)+cot(x+arctan4))+"c"`

Explanation:

We want to find `int1/(sinx+4cosx)dx`

First we combine the denominator using the Harmonic Addition Theorem

`asinx+bcosx=Rsin(x+alpha)` where `R=sqrt(a^2+b^2)` and `alpha=arctan(b"/"a)`

`sinx+4cosx=sqrt17 sin(x+arctan4)`

So

`int1/(sinx+4cosx)dx=int1/(sqrt17sin(x+arctan4))=1/sqrt17int csc(x+arctan4)dx`

We now use the standard result

`intcscthetad theta=-ln(csctheta+cottheta)`

So

`1/sqrt17int csc (x+ arctan4)dx=-1/sqrt17ln(csc(x+arctan4)+cot(x+arctan4))+"c"`