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

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

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integral of sin^5(θ)cos^5(θ)

\int sin^{5} (θ) cos^{5} (θ) d θ

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

Solution steps

\int sin^{5} (θ) cos^{5} (θ) d θ

Rewrite using trig identities

= \int (1 - cos^{2} (θ))^{2} sin (θ) cos^{5} (θ) d θ

Apply u-substitution

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

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

Expand

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

= \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

= - \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}

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

Substitute back

u = cos (θ)

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

Add a constant to the solution

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

x + 2, - 3 x + 6, 0, 1

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f (x) = cos (x) + 7

\frac{d}{d x} (arctan (2^{x}))

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