What is the integral of sin^5(x)cos^6(x) ?

The integral of sin^5(x)cos^6(x) is -(cos^7(x))/7+(2cos^9(x))/9-(cos^{11}(x))/(11)+C

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integral of sin^5(x)cos^6(x)

\int sin^{5} (x) cos^{6} (x) d x

- \frac{cos ^{7} ( x )}{7} + \frac{2 cos ^{9} ( x )}{9} - \frac{cos ^{11} ( x )}{11} + C

Solution steps

\int sin^{5} (x) cos^{6} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x))^{2} sin (x) cos^{6} (x) d x

Apply u-substitution

: \int - u^{6} (1 - u^{2})^{2} d u

= \int - u^{6} (1 - u^{2})^{2} d u

Expand

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

= \int - u^{6} + 2 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^{6} d u + \int 2 u^{8} d u - \int u^{10} d u

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

\int 2 u^{8} d u = \frac{2 u ^{9}}{9}

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

= - \frac{u ^{7}}{7} + \frac{2 u ^{9}}{9} - \frac{u ^{11}}{11}

Substitute back

u = cos (x)

= - \frac{cos ^{7} ( x )}{7} + \frac{2 cos ^{9} ( x )}{9} - \frac{cos ^{11} ( x )}{11}

Add a constant to the solution

= - \frac{cos ^{7} ( x )}{7} + \frac{2 cos ^{9} ( x )}{9} - \frac{cos ^{11} ( x )}{11} + C

\frac{d}{d x} (x^{3} - 12 x + 3)

\frac{d y}{d t} = \frac{1}{t y + t + y + 1}

\int_{0}^{1} \frac{2 x}{( x ^{2} + 4 ) ^{3}} d x

4 - x, 0, 4

What is the integral of sin^5(x)cos^6(x) ?
The integral of sin^5(x)cos^6(x) is -(cos^7(x))/7+(2cos^9(x))/9-(cos^{11}(x))/(11)+C