What is the integral of sin^3(x)cos^{16}(x) ?

The integral of sin^3(x)cos^{16}(x) is -(cos^{17}(x))/(17)+(cos^{19}(x))/(19)+C

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integral of sin^3(x)cos^{16}(x)

\int sin^{3} (x) cos^{16} (x) d x

- \frac{cos ^{17} ( x )}{17} + \frac{cos ^{19} ( x )}{19} + C

Solution steps

\int sin^{3} (x) cos^{16} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{16} (x) d x

Apply u-substitution

: \int - u^{16} (1 - u^{2}) d u

= \int - u^{16} (1 - u^{2}) d u

Expand

- u^{16} (1 - u^{2}) : - u^{16} + u^{18}

= \int - u^{16} + u^{18} 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^{16} d u + \int u^{18} d u

\int u^{16} d u = \frac{u ^{17}}{17}

\int u^{18} d u = \frac{u ^{19}}{19}

= - \frac{u ^{17}}{17} + \frac{u ^{19}}{19}

Substitute back

u = cos (x)

= - \frac{cos ^{17} ( x )}{17} + \frac{cos ^{19} ( x )}{19}

Add a constant to the solution

= - \frac{cos ^{17} ( x )}{17} + \frac{cos ^{19} ( x )}{19} + C

y \to 0 lim (\frac{x ^{y} - 1}{y})

8 8 x^{2} + 5

x \to 0 - lim (\frac{4}{sin ( x )})

\int x e^{- a x^{2}} d x

y^{^{'}} = 4 e^{x + y}

What is the integral of sin^3(x)cos^{16}(x) ?
The integral of sin^3(x)cos^{16}(x) is -(cos^{17}(x))/(17)+(cos^{19}(x))/(19)+C