What is the integral of (sin^3(x))/(cos(x)) ?

The integral of (sin^3(x))/(cos(x)) is -ln|cos(x)|+(cos^2(x))/2+C

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integral of (sin^3(x))/(cos(x))

integral

\frac{sin ^{3} ( x )}{cos ( x )}

- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2} + C

Solution steps

\int \frac{sin ^{3} ( x )}{cos ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

= \int - \frac{1 - u ^{2}}{u} d u

Expand

- \frac{1 - u ^{2}}{u} : - \frac{1}{u} + u

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

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}{u} d u + \int u d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= - ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = cos (x)

= - ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2}

Add a constant to the solution

= - ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2} + C

in t e g r a l \frac{x ^{2}}{x ^{2} + 5}

in t e g r a l \frac{( e ^{x} ) ^{2}}{x}

in t e g r a l - x^{2} ln (x)

in t e g r a l \frac{x ^{2}}{x ^{2} + 1}

in t e g r a l y (4 - y)^{\frac{1}{2}}

What is the integral of (sin^3(x))/(cos(x)) ?
The integral of (sin^3(x))/(cos(x)) is -ln|cos(x)|+(cos^2(x))/2+C