What is the integral of tan(y),y ?

The integral of tan(y),y is -ln|cos(y)|+C

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integral of tan(y) with respect to y

integral

tan (y), y

- ln ∣ cos (y) ∣ + C

Solution steps

\int tan (y) d y

Rewrite using trig identities

= \int \frac{sin ( y )}{cos ( y )} d y

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = cos (y)

= - ln ∣ cos (y) ∣

Add a constant to the solution

= - ln ∣ cos (y) ∣ + C

- \frac{d}{d t} (cos (2 t))

- \frac{d}{d x} (2 cos (2 x))

- \frac{d}{d x} (\frac{1}{( e ^{x} )})

\frac{d y}{d x} y = x sin (x)

\frac{1}{( y ^{2} )}