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

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

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integral of 5cot^5(θ)sin^4(θ)

\int 5 cot^{5} (θ) sin^{4} (θ) d θ

5 (ln ∣ sin (θ) ∣ - sin^{2} (θ) + \frac{sin ^{4} ( θ )}{4}) + C

Solution steps

\int 5 cot^{5} (θ) sin^{4} (θ) d θ

Take the constant out:

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

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

Rewrite using trig identities

= 5 \cdot \int \frac{( 1 - sin ^{2} ( θ ) ) ^{2} cos ( θ )}{sin ( θ )} d θ

Apply u-substitution

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

= 5 \cdot \int \frac{( 1 - u ^{2} ) ^{2}}{u} d u

Expand

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

= 5 \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

= 5 (\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}

= 5 (ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4})

Substitute back

u = sin (θ)

= 5 (ln ∣ sin (θ) ∣ - sin^{2} (θ) + \frac{sin ^{4} ( θ )}{4})

Add a constant to the solution

= 5 (ln ∣ sin (θ) ∣ - sin^{2} (θ) + \frac{sin ^{4} ( θ )}{4}) + C

9 x

\frac{\partial}{\partial x} (lo g_{10} (x + 1))

x \to 4 lim (\frac{5 x}{x - 4})

f (x) = x^{2} - 9 x, at P (x) = 3, - 15

y = x^{4} + 8 e^{x}, (0, 8)

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