What is the integral of 12sin^5(x)cos^5(x) ?

The integral of 12sin^5(x)cos^5(x) is 12(-(cos^6(x))/6+(cos^8(x))/4-(cos^{10}(x))/(10))+C

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integral of 12sin^5(x)cos^5(x)

\int 12 sin^{5} (x) cos^{5} (x) d x

12 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{4} - \frac{cos ^{10} ( x )}{10}) + C

Solution steps

\int 12 sin^{5} (x) cos^{5} (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^{5} (x) cos^{5} (x) d x

Rewrite using trig identities

= 12 \cdot \int (1 - cos^{2} (x))^{2} sin (x) cos^{5} (x) d x

Apply u-substitution

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

= 12 \cdot \int - u^{5} (1 - u^{2})^{2} d u

Expand

- u^{5} (1 - u^{2})^{2} : - u^{5} + 2 u^{7} - u^{9}

= 12 \cdot \int - u^{5} + 2 u^{7} - u^{9} 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^{5} d u + \int 2 u^{7} d u - \int u^{9} d u)

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

\int 2 u^{7} d u = \frac{u ^{8}}{4}

\int u^{9} d u = \frac{u ^{10}}{10}

= 12 (- \frac{u ^{6}}{6} + \frac{u ^{8}}{4} - \frac{u ^{10}}{10})

Substitute back

u = cos (x)

= 12 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{4} - \frac{cos ^{10} ( x )}{10})

Add a constant to the solution

= 12 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{4} - \frac{cos ^{10} ( x )}{10}) + C

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What is the integral of 12sin^5(x)cos^5(x) ?
The integral of 12sin^5(x)cos^5(x) is 12(-(cos^6(x))/6+(cos^8(x))/4-(cos^{10}(x))/(10))+C