What is the integral of 3cot^5(θ)sin^4(θ) ?

The integral of 3cot^5(θ)sin^4(θ) is 3(ln|sin(θ)|-sin^2(θ)+(sin^4(θ))/4)+C

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integral of 3cot^5(θ)sin^4(θ)

\int 3 cot^{5} (θ) sin^{4} (θ) d θ

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

Solution steps

\int 3 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

= 3 \cdot \int cot^{5} (θ) sin^{4} (θ) d θ

Rewrite using trig identities

= 3 \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

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

Expand

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

= 3 \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

= 3 (\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}

= 3 (ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4})

Substitute back

u = sin (θ)

= 3 (ln ∣ sin (θ) ∣ - sin^{2} (θ) + \frac{sin ^{4} ( θ )}{4})

Add a constant to the solution

= 3 (ln ∣ sin (θ) ∣ - sin^{2} (θ) + \frac{sin ^{4} ( θ )}{4}) + C

x^{2} - 2 x + 1

arctan (x - \frac{π}{2})

\int x^{3} x^{2} + 35 d x

e^{(x^{3})}

What is the integral of 3cot^5(θ)sin^4(θ) ?
The integral of 3cot^5(θ)sin^4(θ) is 3(ln|sin(θ)|-sin^2(θ)+(sin^4(θ))/4)+C