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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

= \int \frac{1}{u} + \frac{- u + 1}{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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{7}{x ^{3}} d x

y = x^{10} e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = t^{2}

2 x y d x - (1 + y) d y = 0

\int \frac{3 x}{5 x ^{2} - 1} d x

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