What is the integral of (sec^2(y))/(tan(y)) ?

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

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integral of (sec^2(y))/(tan(y))

\int \frac{sec ^{2} ( y )}{tan ( y )} d y

ln ∣ tan (y) ∣ + C

Solution steps

\int \frac{sec ^{2} ( y )}{tan ( y )} d y

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = tan (y)

= ln ∣ tan (y) ∣

Add a constant to the solution

= ln ∣ tan (y) ∣ + C

\frac{d}{d x} (sin^{2} (tan (4 x)))

\int tan (θ) sec (θ) d θ

y^{^{'}} = e^{4 x + y}

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

x \to 2 - lim (\frac{1}{( x - 2 ) ^{2}})

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