What is the integral of sin^7(x)cos^3(x) ?

The integral of sin^7(x)cos^3(x) is -(cos^4(x))/4+(cos^6(x))/2-(3cos^8(x))/8+(cos^{10}(x))/(10)+C

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integral of sin^7(x)cos^3(x)

\int sin^{7} (x) cos^{3} (x) d x

- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{2} - \frac{3 cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10} + C

Solution steps

\int sin^{7} (x) cos^{3} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x))^{3} sin (x) cos^{3} (x) d x

Apply u-substitution

: \int - u^{3} (1 - u^{2})^{3} d u

= \int - u^{3} (1 - u^{2})^{3} d u

Expand

- u^{3} (1 - u^{2})^{3} : - u^{3} + 3 u^{5} - 3 u^{7} + u^{9}

= \int - u^{3} + 3 u^{5} - 3 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^{3} d u + \int 3 u^{5} d u - \int 3 u^{7} d u + \int u^{9} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int 3 u^{5} d u = \frac{u ^{6}}{2}

\int 3 u^{7} d u = \frac{3 u ^{8}}{8}

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

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{2} - \frac{3 u ^{8}}{8} + \frac{u ^{10}}{10}

Substitute back

u = cos (x)

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{2} - \frac{3 cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}

Add a constant to the solution

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{2} - \frac{3 cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10} + C

x \to 0 lim (x^{19 x})

\int_{1}^{\infty} f (x) d x

(19 x y^{2} - 15 y) + (19 x^{2} y - 15 x) y^{^{'}} = 0

\int \frac{sec ( x ) + cos ( x )}{7 cos ( x )} d x

\frac{d}{d x} (x^{4} \frac{1}{8} + \frac{1}{4 x ^{2}})

What is the integral of sin^7(x)cos^3(x) ?
The integral of sin^7(x)cos^3(x) is -(cos^4(x))/4+(cos^6(x))/2-(3cos^8(x))/8+(cos^{10}(x))/(10)+C