What is the integral of sin^3(θ)cos^8(θ) ?

The integral of sin^3(θ)cos^8(θ) is -(cos^9(θ))/9+(cos^{11}(θ))/(11)+C

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integral of sin^3(θ)cos^8(θ)

\int sin^{3} (θ) cos^{8} (θ) d θ

- \frac{cos ^{9} ( θ )}{9} + \frac{cos ^{11} ( θ )}{11} + C

Solution steps

\int sin^{3} (θ) cos^{8} (θ) d θ

Rewrite using trig identities

= \int (1 - cos^{2} (θ)) sin (θ) cos^{8} (θ) d θ

Apply u-substitution

: \int - u^{8} (1 - u^{2}) d u

= \int - u^{8} (1 - u^{2}) d u

Expand

- u^{8} (1 - u^{2}) : - u^{8} + u^{10}

= \int - u^{8} + u^{10} 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 u^{8} d u + \int u^{10} d u

\int u^{8} d u = \frac{u ^{9}}{9}

\int u^{10} d u = \frac{u ^{11}}{11}

= - \frac{u ^{9}}{9} + \frac{u ^{11}}{11}

Substitute back

u = cos (θ)

= - \frac{cos ^{9} ( θ )}{9} + \frac{cos ^{11} ( θ )}{11}

Add a constant to the solution

= - \frac{cos ^{9} ( θ )}{9} + \frac{cos ^{11} ( θ )}{11} + C

t \to 0 lim (\frac{e ^{t} - 1}{t ^{3}})

\frac{d}{d x} (2 x^{3} + 3 x^{2} - 12 x)

5 x^{2} - 6 x + 1

f (x) = 9 x sin (x)

f (x) = e^{- 5 x}, at x = 0

What is the integral of sin^3(θ)cos^8(θ) ?
The integral of sin^3(θ)cos^8(θ) is -(cos^9(θ))/9+(cos^{11}(θ))/(11)+C