What is the integral of 9sin^5(x)cos^3(x) ?

The integral of 9sin^5(x)cos^3(x) is 9(-(cos^4(x))/4+(cos^6(x))/3-(cos^8(x))/8)+C

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integral of 9sin^5(x)cos^3(x)

\int 9 sin^{5} (x) cos^{3} (x) d x

9 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{3} - \frac{cos ^{8} ( x )}{8}) + C

Solution steps

\int 9 sin^{5} (x) cos^{3} (x) d x

Take the constant out:

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

= 9 \cdot \int sin^{5} (x) cos^{3} (x) d x

Rewrite using trig identities

= 9 \cdot \int (1 - cos^{2} (x))^{2} sin (x) cos^{3} (x) d x

Apply u-substitution

: \int - u^{3} (1 - u^{2})^{2} d u

= 9 \cdot \int - u^{3} (1 - u^{2})^{2} d u

Expand

- u^{3} (1 - u^{2})^{2} : - u^{3} + 2 u^{5} - u^{7}

= 9 \cdot \int - u^{3} + 2 u^{5} - u^{7} 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

= 9 (- \int u^{3} d u + \int 2 u^{5} d u - \int u^{7} d u)

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

\int 2 u^{5} d u = \frac{u ^{6}}{3}

\int u^{7} d u = \frac{u ^{8}}{8}

= 9 (- \frac{u ^{4}}{4} + \frac{u ^{6}}{3} - \frac{u ^{8}}{8})

Substitute back

u = cos (x)

= 9 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{3} - \frac{cos ^{8} ( x )}{8})

Add a constant to the solution

= 9 (- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{3} - \frac{cos ^{8} ( x )}{8}) + C

f (x) = (8 x - x^{2})^{3}

\int sin (5 x) d x

x \to - 1 lim (x^{3 - 4 x^{2} + 1})

4 t - 4 e^{t}

\frac{\partial}{\partial x} (z sin^{2} (x))

What is the integral of 9sin^5(x)cos^3(x) ?
The integral of 9sin^5(x)cos^3(x) is 9(-(cos^4(x))/4+(cos^6(x))/3-(cos^8(x))/8)+C