What is the integral of ((7sin^3(x)))/(cos(x)) ?

The integral of ((7sin^3(x)))/(cos(x)) is 7(-ln|cos(x)|+(cos^2(x))/2)+C

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integral of ((7sin^3(x)))/(cos(x))

\int \frac{( 7 sin ^{3} ( x ) )}{cos ( x )} d x

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

Solution steps

\int \frac{7 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

= 7 \cdot \int \frac{sin ^{3} ( x )}{cos ( x )} d x

Rewrite using trig identities

= 7 \cdot \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ( x )} d x

Apply u-substitution

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

= 7 \cdot \int - \frac{1 - u ^{2}}{u} d u

Expand

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

= 7 \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

= 7 (- \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}

= 7 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = cos (x)

= 7 (- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2})

Add a constant to the solution

= 7 (- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2}) + C

n = 0 \sum \infty n (sin (\frac{1}{n}))

f (x) = cot^{3} (4 x)

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

e^{5 x}, e^{10 x}, x = 1

(x + 2) (10 - 3 x)

What is the integral of ((7sin^3(x)))/(cos(x)) ?
The integral of ((7sin^3(x)))/(cos(x)) is 7(-ln|cos(x)|+(cos^2(x))/2)+C