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

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

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integral of 1/(1-sin(θ)+cos(θ))

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

- ln - tan (\frac{θ}{2}) + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{- u + 1} d u

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

Apply u-substitution

: \int - \frac{1}{v} d v

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

= - ln - tan (\frac{θ}{2}) + 1

Add a constant to the solution

= - ln - tan (\frac{θ}{2}) + 1 + C

\int 3 e^{4 t} d t

\int \frac{4 + 8 x sin ( x )}{2 x} d x

\int tan^{8} (x) cos^{9} (x) d x

y = 1 0^{x}

(\frac{1}{3}) (\frac{d x}{d t}) - 2 e^{- t} = - x

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