What is the integral of cos^3(x)sin^6(x) ?

The integral of cos^3(x)sin^6(x) is (sin^7(x))/7-(sin^9(x))/9+C

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integral of cos^3(x)sin^6(x)

\int cos^{3} (x) sin^{6} (x) d x

\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

Solution steps

\int cos^{3} (x) sin^{6} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int u^{6} - u^{8} 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 u^{8} d u

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

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

= \frac{u ^{7}}{7} - \frac{u ^{9}}{9}

Substitute back

u = sin (x)

= \frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}

Add a constant to the solution

= \frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

y = 8 i n^{x} + c

t \to 0 + lim (arctan (\frac{1}{t}))

\int (4 cos (θ))^{2} d θ

\frac{\partial}{\partial t} (e^{s i n (x + c t)})

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{- \frac{1}{2}}

What is the integral of cos^3(x)sin^6(x) ?
The integral of cos^3(x)sin^6(x) is (sin^7(x))/7-(sin^9(x))/9+C