Integrating Trigonometric Functions with Substitution

[Penultimate]

How can i solve the integral of 1/(sinx + cosx)?

 

[Dick]

I would start by multiplying numerator and denominator by (sin(x)-cos(x)). Does that give you any ideas?

 

[rootX]

General technique:

1/(sinx + cosx)

Use: t = tan (1/2.x)
and find sin(x), cos(x) in terms of t {Using a traingle helps}.

This solves all kind of problems where you ve sin,cos in denominators and your book should have also discussed this somewhere.

Spoiler

You should first prove that sin(t) = 2x / (1+x^2) and cos(t) = (1-x^2)/(1+x^2)
and dt = 2/(1+x^2).dx

 

[Penultimate]

So the general tecnique is to have t= tan 1/2 of the variable?

 

[Dick]

So the general tecnique is to have t= tan 1/2 of the variable?

It's a general way to replace sin(x), cos(x) and dx with rational functions (quotients of polynomials) of another variable. It should work. It's not necessarily the easiest way to do any given integral.

 

[rootX]

So the general tecnique is to have t= tan 1/2 of the variable?

Just use it when you can't think of anything else. Mostly likely, it becomes a problem of partial fractions and inverse of sines,cosines,tan (integration table always have their integration formulas),.. it's just messing up algebra/using wrong trigonometry relations that make this harder to use but you always know what to do in the next step.

There are too many ways to solve these problems. Here's another way:
convert sin (x) + cos(x) = A.sin(x+phi) .. and you will have 1/A*sin (...), and now you can use integration table.

 

[HallsofIvy]

And don't forget what Dick originally suggested. Multiplying both numerator and denominator of 1/(sin(x)+ cos(x)) by sin(x)- cos(x) you have (sin(x)- cos(x))/(sin²(x)- cos²(x))= sin(x)/(sin²(x)- cos²(x))- cos(x)/(sin²(x)- cos²(x))= sin(x)/(sin²-(1- sin²(x))- cos(x)/((1- cos²(x))- cos²(x))= sin(x)/(2sin²-1)- cos(x)/(1- 2cos²(x)).