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

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

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

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

\frac{sin ^{5} ( y )}{5} - \frac{3 sin ^{7} ( y )}{7} + \frac{sin ^{9} ( y )}{3} - \frac{sin ^{11} ( y )}{11} + C

Solution steps

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

Rewrite using trig identities

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

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 = sin (y)

= \frac{sin ^{5} ( y )}{5} - \frac{3 sin ^{7} ( y )}{7} + \frac{sin ^{9} ( y )}{3} - \frac{sin ^{11} ( y )}{11}

Add a constant to the solution

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

\frac{d ^{2} y}{d x ^{2}}, x^{2} + y^{2} = 36

t \frac{d y}{d t} + 27 y = \frac{1}{t}

\frac{d}{d x} (\frac{2 d}{x ^{3}} - a)

f (x) = \frac{10 x + x ^{2}}{( 5 + x ) ^{2}}

\frac{d y}{d x} - \frac{2 x ^{3}}{y} = 0

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