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

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

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

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

- cos (x) + \frac{sin ^{2} ( x )}{2} + C

Solution steps

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

Expand

\frac{tan ( x ) + sin ( x )}{sec ( x )} : \frac{tan ( x )}{sec ( x )} + \frac{sin ( x )}{sec ( x )}

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

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

\int \frac{tan ( x )}{sec ( x )} d x = - cos (x)

\int \frac{sin ( x )}{sec ( x )} d x = \frac{sin ^{2} ( x )}{2}

= - cos (x) + \frac{sin ^{2} ( x )}{2}

Add a constant to the solution

= - cos (x) + \frac{sin ^{2} ( x )}{2} + C

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\frac{d}{d x} (ln (a + x + x^{2} + 2 a x))

\frac{\partial}{\partial x} (- 2 x y z^{2})

\int_{0}^{1} \frac{1}{x ^{1}} d x

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