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

The integral of (tan(x)sec^2(y)) is -sec^2(y)ln|cos(x)|+C

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

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

- sec^{2} (y) ln ∣ cos (x) ∣ + C

Solution steps

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

Take the constant out:

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

= sec^{2} (y) \cdot \int tan (x) d x

Rewrite using trig identities

= sec^{2} (y) \cdot \int \frac{sin ( x )}{cos ( x )} d x

Apply u-substitution

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

= sec^{2} (y) \cdot \int - \frac{1}{u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= sec^{2} (y) (- ln ∣ u ∣)

Substitute back

u = cos (x)

= sec^{2} (y) (- ln ∣ cos (x) ∣)

Simplify

= - sec^{2} (y) ln ∣ cos (x) ∣

Add a constant to the solution

= - sec^{2} (y) ln ∣ cos (x) ∣ + C

\frac{d}{d x} ((a + b x)^{3})

x \to 2 lim (x^{2} - 4 x + 1)

\int \frac{1}{( 1 + x ^{4} )} d x

\int \frac{x ^{4}}{( 7 + x ^{5} ) ^{2}} d x

n = 2 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

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