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

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

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integral of 1/(tan(x)-sin(x))

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{( u + 1 ) ( u - 1 )}{2 u ^{3}} d u

= \int - \frac{( u + 1 ) ( u - 1 )}{2 u ^{3}} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{( u + 1 ) ( u - 1 )}{u ^{3}} d u

Apply Integration By Parts

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

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

\int - \frac{1}{u} d u = - ln ∣ u ∣

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

4 x \cdot 2^{x}

\frac{d y}{d x} = 20 - \frac{y}{6000 + 20 x}

\int_{0}^{π} cos^{2} (2 x) d x

y^{^{''}} + 64 y = 0

\frac{d}{d x} (5 e^{2})

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