What is the integral of 7cot^5(x)sin^4(x) ?

The integral of 7cot^5(x)sin^4(x) is 7(ln|sin(x)|-sin^2(x)+(sin^4(x))/4)+C

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integral of 7cot^5(x)sin^4(x)

\int 7 cot^{5} (x) sin^{4} (x) d x

7 (ln ∣ sin (x) ∣ - sin^{2} (x) + \frac{sin ^{4} ( x )}{4}) + C

Solution steps

\int 7 cot^{5} (x) sin^{4} (x) d x

Take the constant out:

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

= 7 \cdot \int cot^{5} (x) sin^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{( 1 - u ^{2} ) ^{2}}{u} : \frac{1}{u} - 2 u + u^{3}

= 7 \cdot \int \frac{1}{u} - 2 u + u^{3} 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 2 u d u + \int u^{3} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int u^{3} d u = \frac{u ^{4}}{4}

= 7 (ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4})

Substitute back

u = sin (x)

= 7 (ln ∣ sin (x) ∣ - sin^{2} (x) + \frac{sin ^{4} ( x )}{4})

Add a constant to the solution

= 7 (ln ∣ sin (x) ∣ - sin^{2} (x) + \frac{sin ^{4} ( x )}{4}) + C

\int_{0}^{1} \frac{36}{( 2 x + 1 ) ^{3}} d x

\int \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}} d x

\int sin (2 x) sin (7 x) d x

\frac{d y}{d x} = \frac{sin ( x )}{y}, y (0) = 1

\frac{\partial}{\partial x} (y \cdot e^{- x})

What is the integral of 7cot^5(x)sin^4(x) ?
The integral of 7cot^5(x)sin^4(x) is 7(ln|sin(x)|-sin^2(x)+(sin^4(x))/4)+C