What is the integral of sin^6(5x) ?

The integral of sin^6(5x) is -1/30 sin^5(5x)cos(5x)+1/96 (-4sin^3(5x)cos(5x)+3(10x-sin(10x)))+C

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integral of sin^6(5x)

\int sin^{6} (5 x) d x

- \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x))) + C

Solution steps

\int sin^{6} (5 x) d x

Apply u-substitution

= \int sin^{6} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sin^{6} (u) d u

Apply Integral Reduction

= \frac{1}{5} (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} \cdot \int sin^{4} (u) d u)

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

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

Substitute back

u = 5 x

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

Simplify

\frac{1}{5} (- \frac{cos ( 5 x ) sin ^{5} ( 5 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (2 \cdot 5 x)))) : - \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x)))

= - \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x)))

Add a constant to the solution

= - \frac{1}{30} sin^{5} (5 x) cos (5 x) + \frac{1}{96} (- 4 sin^{3} (5 x) cos (5 x) + 3 (10 x - sin (10 x))) + C

What is the integral of sin^6(5x) ?
The integral of sin^6(5x) is -1/30 sin^5(5x)cos(5x)+1/96 (-4sin^3(5x)cos(5x)+3(10x-sin(10x)))+C