What is the integral of sin^3(10x) ?

The integral of sin^3(10x) is 1/10 (-cos(10x)+(cos^3(10x))/3)+C

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integral of sin^3(10x)

\int sin^{3} (10 x) d x

\frac{1}{10} (- cos (10 x) + \frac{cos ^{3} ( 10 x )}{3}) + C

Solution steps

\int sin^{3} (10 x) d x

Apply u-substitution

= \int sin^{3} (u) \frac{1}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{10} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

= \frac{1}{10} \cdot \int - 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{10} (- \int 1 d v + \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{10} (- v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{10} (- cos (10 x) + \frac{cos ^{3} ( 10 x )}{3})

Add a constant to the solution

= \frac{1}{10} (- cos (10 x) + \frac{cos ^{3} ( 10 x )}{3}) + C

t an g e n t y = x^{2} - x + 4, at x = 4

x \to - 0 lim (\frac{1}{1 + e ^{\frac{1}{x}}})

x \to - 6 lim (x - 10)

\int \frac{2 x ^{4}}{3 - 4 x ^{5}} d x

t an g e n t f (x) = x^{3} - 5 x^{2} - 2 x + 1, at x = - 2

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