What is the integral of 12sin^6(x)cos^3(x) ?

The integral of 12sin^6(x)cos^3(x) is 12((sin^7(x))/7-(sin^9(x))/9)+C

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integral of 12sin^6(x)cos^3(x)

\int 12 sin^{6} (x) cos^{3} (x) d x

12 (\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}) + C

Solution steps

\int 12 sin^{6} (x) cos^{3} (x) d x

Take the constant out:

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

= 12 \cdot \int sin^{6} (x) cos^{3} (x) d x

Rewrite using trig identities

= 12 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{6} (x) d x

Apply u-substitution

: \int u^{6} (1 - u^{2}) d u

= 12 \cdot \int u^{6} (1 - u^{2}) d u

Expand

u^{6} (1 - u^{2}) : u^{6} - u^{8}

= 12 \cdot \int u^{6} - u^{8} 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

= 12 (\int u^{6} d u - \int u^{8} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

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

= 12 (\frac{u ^{7}}{7} - \frac{u ^{9}}{9})

Substitute back

u = sin (x)

= 12 (\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9})

Add a constant to the solution

= 12 (\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}) + C

f (x) = x^{\frac{5}{11}} (x - 5)^{2}

\frac{\partial}{\partial x} (\frac{3 x + 2 y}{y ^{2} + 5 x ^{2}})

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\int \frac{e ^{5 x}}{5} d x

What is the integral of 12sin^6(x)cos^3(x) ?
The integral of 12sin^6(x)cos^3(x) is 12((sin^7(x))/7-(sin^9(x))/9)+C