What is the integral of sin^3(x)cos^4(x) ?

The integral of sin^3(x)cos^4(x) is -(cos^5(x))/5+(cos^7(x))/7+C

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integral of sin^3(x)cos^4(x)

\int sin^{3} (x) cos^{4} (x) d x

- \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7} + C

Solution steps

\int sin^{3} (x) cos^{4} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{4} (x) d x

Apply u-substitution

: \int - u^{4} (1 - u^{2}) d u

= \int - u^{4} (1 - u^{2}) d u

Expand

- u^{4} (1 - u^{2}) : - u^{4} + u^{6}

= \int - u^{4} + u^{6} 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^{4} d u + \int u^{6} d u

\int u^{4} d u = \frac{u ^{5}}{5}

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

= - \frac{u ^{5}}{5} + \frac{u ^{7}}{7}

Substitute back

u = cos (x)

= - \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7}

Add a constant to the solution

= - \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7} + C

\frac{d}{d x} (\frac{x sin ( x )}{x ^{2} - 1})

\int p (p + 4)^{5} d p

\frac{d}{d x} (\frac{e ^{x}}{5 + e ^{x}})

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

\int (1 + e^{x} y + x e^{x} y) d x

What is the integral of sin^3(x)cos^4(x) ?
The integral of sin^3(x)cos^4(x) is -(cos^5(x))/5+(cos^7(x))/7+C