# Integral (Sin[3x+5])/(1+Cos[3x+5]) Solved

Integral (Sin[3x+5])/(1+Cos[3x+5]) solved step by step.

$\bg_white \small \color{black} \int {{\sin (3x+5)} \over {1+\cos (3x+5)}}dx$
First applied substitute method, using the function arguments as substitution
$\bg_white \small \color{black} u={3x+5} \:\: and \:\: du={3dx} \therefore {du \over 3}=dx$
Now,we replace the function argument per u, that is our new argument, and we factorized per 1/3 because du/3=dx
$\bg_white \small \color{black} {1\over3}{\int {{\sin (u)} \over {1+\cos (u)}}du}$
till can’t we directly solve, but we can applied again the substitution method, but now using w as the substitute variable.
$\bg_white \small \color{black} w={\cos(u)+1} \:\: and \:\: dw=-\sin(u)du \therefore -dw=\sin(u)du$
And now, we replace w in the function

$\bg_white \small \color{black} -{1\over3}{\int {1\over {1+w}}dw}$
the latest, we know that integral of 1/x is equal to log(x)
$\bg_white \small \color{black} -{1\over3}{\log(w)+constant}$
now, replace back from w to cos(u)+1
$\bg_white \small \color{black} -{1\over3}{\log(\cos(u)+1)+constant}$
after, replace back from u to 3x+5
$\bg_white \small \color{black} -{1\over3}{\log(\cos(3x+5)+1)+constant}$
finally, we can say that is integral is
$\bg_white \small \color{black} \int {{\sin (3x+5)} \over {1+\cos (3x+5)}}dx=-{1\over3}{\log(\cos(3x+5)+1)+constant}$

Identities used for solved this exercise.
Substitute method
This method say that the integral will can by for a portion of the original integral by the derivative of this same portion.
If we have
$\bg_white \small \color{black} \int {(ab)}dx$
and b=a’
we can say that b=u and du=a’ therefore
$\bg_white \small \color{black} {\int {(ab)}dx} = {\int {udu}}$

Logarithm integral identity
$\bg_white \small \color{black} {\int {1/x}dx} = log(x) + constant$