What is the integral of (sin(x))/(sin(x)+cos(x)+1) ?

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

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integral of (sin(x))/(sin(x)+cos(x)+1)

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

- ln tan (\frac{x}{2}) + 1 + \frac{1}{2} ln (sec^{2} (\frac{x}{2})) + arctan (tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

= \int \frac{2 u}{( u + 1 ) ( 1 + u ^{2} )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u}{( u + 1 ) ( 1 + u ^{2} )} d u

Take the partial fraction of

\frac{u}{( u + 1 ) ( 1 + u ^{2} )} : - \frac{1}{2 ( u + 1 )} + \frac{u + 1}{2 ( u ^{2} + 1 )}

= 2 \cdot \int - \frac{1}{2 ( u + 1 )} + \frac{u + 1}{2 ( u ^{2} + 1 )} d u

Apply the Sum Rule:

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

= 2 (- \int \frac{1}{2 ( u + 1 )} d u + \int \frac{u + 1}{2 ( u ^{2} + 1 )} d u)

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

\int \frac{u + 1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (\frac{1}{2} ln u^{2} + 1 + arctan (u))

= 2 (- \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln u^{2} + 1 + arctan (u)))

Substitute back

u = tan (\frac{x}{2})

= 2 (- \frac{1}{2} ln tan (\frac{x}{2}) + 1 + \frac{1}{2} (\frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1 + arctan (tan (\frac{x}{2}))))

Simplify

2 (- \frac{1}{2} ln tan (\frac{x}{2}) + 1 + \frac{1}{2} (\frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1 + arctan (tan (\frac{x}{2})))) : - ln tan (\frac{x}{2}) + 1 + \frac{1}{2} ln (sec^{2} (\frac{x}{2})) + arctan (tan (\frac{x}{2}))

= - ln tan (\frac{x}{2}) + 1 + \frac{1}{2} ln (sec^{2} (\frac{x}{2})) + arctan (tan (\frac{x}{2}))

Add a constant to the solution

= - ln tan (\frac{x}{2}) + 1 + \frac{1}{2} ln (sec^{2} (\frac{x}{2})) + arctan (tan (\frac{x}{2})) + C

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