What is the integral of-tan(y) ?

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

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integral of-tan(y)

\int - tan (y) d y

ln ∣ cos (y) ∣ + C

Solution steps

\int - tan (y) d y

Take the constant out:

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

= - \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) ∣)

Simplify

= ln ∣ cos (y) ∣

Add a constant to the solution

= ln ∣ cos (y) ∣ + C

\int \frac{1}{5 - 4 cos ( x )} d x

t y^{^{'}} = t + y

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

\frac{( s + 5 )}{s ( s + 4 ) ^{2}}

\int \frac{cos ( x )}{5 + sin ^{2} ( x )} d x