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

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

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integral of cot^5(θ)sin^4(θ)

\int cot^{5} (θ) sin^{4} (θ) d θ

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

Solution steps

\int cot^{5} (θ) sin^{4} (θ) d θ

Rewrite using trig identities

= \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

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

Expand

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

= \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

= \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}

= ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4}

Substitute back

u = sin (θ)

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

Add a constant to the solution

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

x \to 0 lim (x^{6} cos (\frac{7}{x}))

\frac{d}{d x} ((16 c x - 4 x^{2})^{\frac{1}{2}})

x \to 0 lim (x^{4} - \frac{x ^{3}}{3})

y = 3 x^{2}, y = 18 x - 27, [0, 3]

\int sin^{2} (x) cos (z) d z

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