What is the integral of (sin(x))/(cos(x)+sin(x)) ?

The integral of (sin(x))/(cos(x)+sin(x)) is 1/2 x-1/2 ln|cos(x)+sin(x)|+C

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integral of (sin(x))/(cos(x)+sin(x))

\int \frac{sin ( x )}{cos ( x ) + sin ( x )} d x

\frac{1}{2} x - \frac{1}{2} ln ∣ cos (x) + sin (x) ∣ + C

Solution steps

\int \frac{sin ( x )}{cos ( x ) + sin ( x )} d x

Rewrite using trig identities

= \int \frac{1}{2} + \frac{sin ( x ) - cos ( x )}{2 ( cos ( x ) + sin ( x ) )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2} d x + \int \frac{sin ( x ) - cos ( x )}{2 ( cos ( x ) + sin ( x ) )} d x

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{sin ( x ) - cos ( x )}{2 ( cos ( x ) + sin ( x ) )} d x = - \frac{1}{2} ln ∣ cos (x) + sin (x) ∣

= \frac{1}{2} x - \frac{1}{2} ln ∣ cos (x) + sin (x) ∣

Add a constant to the solution

= \frac{1}{2} x - \frac{1}{2} ln ∣ cos (x) + sin (x) ∣ + C

\int \frac{e ^{8 x}}{e ^{16 x} + 16} d x

\int_{1}^{4} \frac{6}{x ^{2}} d x

(x - 1)^{2} e^{x}

\frac{d}{d x} ((2 x - 1)^{2} ln (2 x - 1))

f (x) = (x) (3 x - 2)

What is the integral of (sin(x))/(cos(x)+sin(x)) ?
The integral of (sin(x))/(cos(x)+sin(x)) is 1/2 x-1/2 ln|cos(x)+sin(x)|+C