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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= 9 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{5} (x) d x

Apply u-substitution

: \int - u^{5} (1 - u^{2}) d u

= 9 \cdot \int - u^{5} (1 - u^{2}) d u

Expand

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

= 9 \cdot \int - 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^{5} d u + \int u^{7} d u)

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

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

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

Substitute back

u = cos (x)

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

Add a constant to the solution

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

f (x) = e^{x} - 7 x^{6}

\int csc (θ) + 3 d θ

\frac{d y}{d x} + 9 y - 4 e^{- 2 x} = 0

3 1 + 5 x

\int \frac{1}{( 4 x ^{2} - 24 x + 27 ) ^{\frac{3}{2}}} d x

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