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

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

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

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

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

Solution steps

\int \frac{1}{sin ( x ) + tan ( 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

\frac{d}{d x} (i sin (x))

\frac{d}{d x} (arccot (10 x))

y^{^{''}} = (y^{^{'}})^{2}

\int_{0}^{l n (9)} x e^{x} d x

x (\frac{d y}{d x}) + y = e^{x}, y (1) = 2

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