What is the integral of sin^7(x)cos^4(x) ?

The integral of sin^7(x)cos^4(x) is -(cos^5(x))/5+(3cos^7(x))/7-(cos^9(x))/3+(cos^{11}(x))/(11)+C

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integral of sin^7(x)cos^4(x)

\int sin^{7} (x) cos^{4} (x) d x

- \frac{cos ^{5} ( x )}{5} + \frac{3 cos ^{7} ( x )}{7} - \frac{cos ^{9} ( x )}{3} + \frac{cos ^{11} ( x )}{11} + C

Solution steps

\int sin^{7} (x) cos^{4} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x))^{3} sin (x) cos^{4} (x) d x

Apply u-substitution

: \int - u^{4} (1 - u^{2})^{3} d u

= \int - u^{4} (1 - u^{2})^{3} d u

Expand

- u^{4} (1 - u^{2})^{3} : - u^{4} + 3 u^{6} - 3 u^{8} + u^{10}

= \int - u^{4} + 3 u^{6} - 3 u^{8} + u^{10} 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 3 u^{6} d u - \int 3 u^{8} d u + \int u^{10} d u

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

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

\int 3 u^{8} d u = \frac{u ^{9}}{3}

\int u^{10} d u = \frac{u ^{11}}{11}

= - \frac{u ^{5}}{5} + \frac{3 u ^{7}}{7} - \frac{u ^{9}}{3} + \frac{u ^{11}}{11}

Substitute back

u = cos (x)

= - \frac{cos ^{5} ( x )}{5} + \frac{3 cos ^{7} ( x )}{7} - \frac{cos ^{9} ( x )}{3} + \frac{cos ^{11} ( x )}{11}

Add a constant to the solution

= - \frac{cos ^{5} ( x )}{5} + \frac{3 cos ^{7} ( x )}{7} - \frac{cos ^{9} ( x )}{3} + \frac{cos ^{11} ( x )}{11} + C

\int - B e^{- 2 t} d t

x^{4} - 2 x^{2} + 3

x \to 7 lim (- 7)

\frac{\partial}{\partial z} (e^{7 yz})

\frac{d}{d x} (x^{3} - 3 x^{2} + 4 x - 1)

What is the integral of sin^7(x)cos^4(x) ?
The integral of sin^7(x)cos^4(x) is -(cos^5(x))/5+(3cos^7(x))/7-(cos^9(x))/3+(cos^{11}(x))/(11)+C