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

The integral of 1/(tan(x)+sin(x)) is -1/2 ((tan^2(x/2))/2-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{tan ^{2} ( \frac{x}{2} )}{2} - 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} d u

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

Expand

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

= - \frac{1}{2} \cdot \int u - \frac{1}{u} 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

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

\int u d u = \frac{u ^{2}}{2}

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

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

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

Add a constant to the solution

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

x \to 1 + lim (\frac{2}{x ^{3} - 1})

\int - 2 π cos^{4} (π x) sin^{5} (π x) d x

\int 5 x 2 x + 3 d x

f (x) = x (3 - x)^{2}, at x = 2

\int \frac{1}{x ( 1 + x ) ^{3}} d x

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