What is the integral of 9sin^6(x) ?

The integral of 9sin^6(x) is 9(-(cos(x)sin^5(x))/6+5/6 (-1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x))))+C

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integral of 9sin^6(x)

\int 9 sin^{6} (x) d x

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

Solution steps

\int 9 sin^{6} (x) d x

Take the constant out:

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

= 9 \cdot \int sin^{6} (x) d x

Apply Integral Reduction

= 9 (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} \cdot \int sin^{4} (x) d x)

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

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

Add a constant to the solution

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

y^{^{'}} + y^{3} x + 9 y = 0

\int x^{5} + 8 x^{- 3} + 2 d x

\frac{d}{d x} (3 x^{3} + 5 x)

\frac{\partial}{\partial x} (x y - ln (z))

\int e^{x + y} d y

What is the integral of 9sin^6(x) ?
The integral of 9sin^6(x) is 9(-(cos(x)sin^5(x))/6+5/6 (-1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x))))+C