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

x \to - 1 lim (\frac{x + 1}{∣ x + 1 ∣})

\int_{1}^{9} \frac{1}{x} d x

x \frac{d y}{d x} + (11 + x) y = e^{- x}

\int \frac{x + 1}{x ^{3} + x} d x

\int \frac{1}{x 1 - x} 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