What is the integral of 4sin^6(x)cos^3(x) ?

The integral of 4sin^6(x)cos^3(x) is 4((sin^7(x))/7-(sin^9(x))/9)+C

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integral of 4sin^6(x)cos^3(x)

\int 4 sin^{6} (x) cos^{3} (x) d x

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

Solution steps

\int 4 sin^{6} (x) cos^{3} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int sin^{6} (x) cos^{3} (x) d x

Rewrite using trig identities

= 4 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{6} (x) d x

Apply u-substitution

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

= 4 \cdot \int u^{6} (1 - u^{2}) d u

Expand

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

= 4 \cdot \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

= 4 (\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}

= 4 (\frac{u ^{7}}{7} - \frac{u ^{9}}{9})

Substitute back

u = sin (x)

= 4 (\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9})

Add a constant to the solution

= 4 (\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}) + C

(y + x^{2} - y^{2}) d x - x d y = 0

\int 4 x^{2} e^{x} d x

f (x) = x^{3} - 6 x^{2} + 12 x - 9

\frac{\partial}{\partial y} (- 0.2 y + 23)

\frac{\partial}{\partial x} (- 5 x y (9 - x - y))

What is the integral of 4sin^6(x)cos^3(x) ?
The integral of 4sin^6(x)cos^3(x) is 4((sin^7(x))/7-(sin^9(x))/9)+C