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

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

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

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

ln ∣ tan (x) + sec (x) ∣ - sin (x) + C

Solution steps

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

Simplify

= \int sin (x) tan (x) d x

Rewrite using trig identities

= \int \frac{1 - cos ^{2} ( x )}{cos ( x )} d x

Expand

\frac{1 - cos ^{2} ( x )}{cos ( x )} : \frac{1}{cos ( x )} - cos (x)

= \int \frac{1}{cos ( x )} - cos (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{1}{cos ( x )} d x - \int cos (x) d x

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

\int cos (x) d x = sin (x)

= ln ∣ tan (x) + sec (x) ∣ - sin (x)

Add a constant to the solution

= ln ∣ tan (x) + sec (x) ∣ - sin (x) + C

\int_{0}^{4} \frac{1}{16 - x ^{2}} d x

x \to 0 + lim (x^{2 x})

\frac{\partial}{\partial t} (e^{t})

n = 2 \sum \infty \frac{x ^{n}}{n}

d y + (3 x^{2} y - x^{2}) d x = 0

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