What is the integral of 1/(sec(y)tan(y)) ?

The integral of 1/(sec(y)tan(y)) is ln|tan(y/2)|+cos(y)+C

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integral of 1/(sec(y)tan(y))

\int \frac{1}{sec ( y ) tan ( y )} d y

ln tan (\frac{y}{2}) + cos (y) + C

Solution steps

\int \frac{1}{sec ( y ) tan ( y )} d y

Rewrite using trig identities

= \int \frac{1 - sin ^{2} ( y )}{sin ( y )} d y

Expand

\frac{1 - sin ^{2} ( y )}{sin ( y )} : \frac{1}{sin ( y )} - sin (y)

= \int \frac{1}{sin ( y )} - sin (y) d y

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}{sin ( y )} d y - \int sin (y) d y

\int \frac{1}{sin ( y )} d y = ln tan (\frac{y}{2})

\int sin (y) d y = - cos (y)

= ln tan (\frac{y}{2}) - (- cos (y))

Simplify

= ln tan (\frac{y}{2}) + cos (y)

Add a constant to the solution

= ln tan (\frac{y}{2}) + cos (y) + C

y^{^{'}} + 2 y = e^{3 x}

\frac{d}{d x} (5 tan (2 x^{3}))

\int b x d x

n = 1 \sum \infty 9 (\frac{1}{10})^{n}

\frac{d}{d x} (x^{c o s (x)})

What is the integral of 1/(sec(y)tan(y)) ?
The integral of 1/(sec(y)tan(y)) is ln|tan(y/2)|+cos(y)+C