Question

Integration, I showed what I did, just need someone to verify that it's right:

Answer 1

\[\LARGE \int\frac{dx}{cosx-sinx+1}\]
\(\Large z=tan\frac{x}{2}\) so \(\Large dx=\frac{2dz}{1+z^2}\)
Sub in:
\[\LARGE \int\frac{\frac{2dz}{1+z^2+1}}{\frac{1-z^2}{1+z^2}-\frac{2z}{1+z^2}+1}\]
Multiply by \(1+z^2\)
\[\LARGE 2\int \frac{dz}{1-z^2-2z+1+z^2}\]
Simplify it
\[\LARGE 2\int \frac{dz}{2-2z}\]
factor and simplfy
\[\LARGE \int \frac{dz}{1-z}\]

\[\LARGE -ln|1-z|+C\]
\[\LARGE -ln|1-tan\frac{x}{2}|+C\]

Answer 2

You lost me at "Luigi0210"

Answer 3

You lost me at Hi

Answer 4

Fine, you lost me at ∫

Answer 5

u=tan(x/2), du = 1/2 sec^2(x/2)dx

\[sinx = \frac{ 2u }{ u^2+1 }\]
\[cosx = \frac{ 1-u^2 }{ u^2+1 }\]
\[dx = \frac{ 2du }{ u^2+1 }\]

\[\huge \int\limits \frac{ 2 }{ (u^2+1)(\frac{ 2u }{ -u^2+1 }+\frac{ 1-u^2 }{ u^2+1 }+1) }du\]

Answer 6

Once again, we are saved by Batman!

Answer 7

\[ = \int\limits \frac{ 1 }{ 1-u }du, ~~~ t = 1-u, dt = -du \implies -\int\limits \frac{ 1 }{ t }dt\]

Answer 8

\[-\ln(1-u)+c \implies -\ln(1-\tan \frac{ x }{ 2 })+C\]

Good job, but there is a better answer if you have restricted x values, not sure if you care for it though lol.