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

The integral of sin^4(y)cos^4(y) is

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

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

- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (y) cos (y) + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)))) - \frac{1}{8} sin^{7} (y) cos (y) + C

Solution steps

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

Rewrite using trig identities

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

Expand

sin^{4} (y) (1 - sin^{2} (y))^{2} : sin^{4} (y) - 2 sin^{6} (y) + sin^{8} (y)

= \int sin^{4} (y) - 2 sin^{6} (y) + sin^{8} (y) d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int sin^{4} (y) d y - \int 2 sin^{6} (y) d y + \int sin^{8} (y) d y

\int sin^{4} (y) d y = - \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y))

\int 2 sin^{6} (y) d y = 2 (- \frac{cos ( y ) sin ^{5} ( y )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y))))

\int sin^{8} (y) d y = - \frac{cos ( y ) sin ^{7} ( y )}{8} + \frac{7}{8} (- \frac{cos ( y ) sin ^{5} ( y )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y))))

= - \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)) - 2 (- \frac{cos ( y ) sin ^{5} ( y )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)))) - \frac{cos ( y ) sin ^{7} ( y )}{8} + \frac{7}{8} (- \frac{cos ( y ) sin ^{5} ( y )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y))))

Simplify

- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)) - 2 (- \frac{cos ( y ) sin ^{5} ( y )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)))) - \frac{cos ( y ) sin ^{7} ( y )}{8} + \frac{7}{8} (- \frac{cos ( y ) sin ^{5} ( y )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)))) : - \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (y) cos (y) + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)))) - \frac{1}{8} sin^{7} (y) cos (y)

= - \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (y) cos (y) + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)))) - \frac{1}{8} sin^{7} (y) cos (y)

Add a constant to the solution

= - \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)) - \frac{9}{8} (- \frac{1}{6} sin^{5} (y) cos (y) + \frac{5}{6} (- \frac{1}{4} sin^{3} (y) cos (y) + \frac{3}{8} (y - \frac{1}{2} sin (2 y)))) - \frac{1}{8} sin^{7} (y) cos (y) + C

f (x) = sin (π x)

x \to 5 lim (5 - x^{2})

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

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 e^{- x} + 3

f (x) = \frac{1}{4 ( 1 + x ^{2} )}, g (x) = \frac{1}{8} x^{2}