What is the integral of ((9sin^3(x)))/(cos(x)) ?

The integral of ((9sin^3(x)))/(cos(x)) is 9(-ln|cos(x)|+(cos^2(x))/2)+C

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integral of ((9sin^3(x)))/(cos(x))

\int \frac{( 9 sin ^{3} ( x ) )}{cos ( x )} d x

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

Solution steps

\int \frac{9 sin ^{3} ( x )}{cos ( x )} d x

Take the constant out:

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

= 9 \cdot \int \frac{sin ^{3} ( x )}{cos ( x )} d x

Rewrite using trig identities

= 9 \cdot \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ( x )} d x

Apply u-substitution

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

= 9 \cdot \int - \frac{1 - u ^{2}}{u} d u

Expand

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

= 9 \cdot \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

= 9 (- \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}

= 9 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = cos (x)

= 9 (- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2})

Add a constant to the solution

= 9 (- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2}) + C

x \to \infty lim (\frac{1}{5 x})

\int (x \cdot sin (x^{2} + 1)) d x

\frac{d}{d x} (e^{\frac{- x ^{2}}{2}})

2 x + \frac{4 x}{3 x - 1}

x \to 3 lim (\frac{2 x + 3}{x - 3})

What is the integral of ((9sin^3(x)))/(cos(x)) ?
The integral of ((9sin^3(x)))/(cos(x)) is 9(-ln|cos(x)|+(cos^2(x))/2)+C