What is the integral of-(sin^2(x))/(cos(x)) ?

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

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integral of-(sin^2(x))/(cos(x))

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

- ln ∣ tan (x) + sec (x) ∣ + sin (x) + C

Solution steps

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

Take the constant out:

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

= - \int \frac{sin ^{2} ( x )}{cos ( x )} d x

Rewrite using trig identities

= - \int \frac{1 - cos ^{2} ( x )}{cos ( x )} d x

Expand

\frac{1 - cos ^{2} ( x )}{cos ( x )} : \frac{1}{cos ( x )} - cos (x)

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

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}{cos ( x )} d x - \int cos (x) d x)

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

\int cos (x) d x = sin (x)

= - (ln ∣ tan (x) + sec (x) ∣ - sin (x))

Simplify

= - ln ∣ tan (x) + sec (x) ∣ + sin (x)

Add a constant to the solution

= - ln ∣ tan (x) + sec (x) ∣ + sin (x) + C

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} - 1})

\frac{sin ( 4 t )}{cos ( 6 t )}

y = x^{3 x}

y^{^{''}} + y = sec (t)

n = 1 \sum \infty \frac{6 n}{9 - 2 n}

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