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

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

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integral of 9sin^6(z)

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Reduction

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

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

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

Add a constant to the solution

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

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x \to 0 lim (cos (3 x))

\int x \cdot ab \cdot e^{- ab x + ab c} d x

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