What is the integral of sin^5(7x) ?

The integral of sin^5(7x) is 1/7 (-cos(7x)+(2cos^3(7x))/3-(cos^5(7x))/5)+C

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integral of sin^5(7x)

\int sin^{5} (7 x) d x

\frac{1}{7} (- cos (7 x) + \frac{2 cos ^{3} ( 7 x )}{3} - \frac{cos ^{5} ( 7 x )}{5}) + C

Solution steps

\int sin^{5} (7 x) d x

Apply u-substitution

: \int sin^{5} (u) \frac{1}{7} d u

= \int sin^{5} (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int sin^{5} (u) d u

Rewrite using trig identities

= \frac{1}{7} \cdot \int (1 - cos^{2} (u))^{2} sin (u) d u

Apply u-substitution

: \int - 1 + 2 v^{2} - v^{4} d v

= \frac{1}{7} \cdot \int - 1 + 2 v^{2} - v^{4} 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}{7} (- \int 1 d v + \int 2 v^{2} d v - \int v^{4} d v)

\int 1 d v = v

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

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{7} (- v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= \frac{1}{7} (- cos (7 x) + \frac{2 cos ^{3} ( 7 x )}{3} - \frac{cos ^{5} ( 7 x )}{5})

Add a constant to the solution

= \frac{1}{7} (- cos (7 x) + \frac{2 cos ^{3} ( 7 x )}{3} - \frac{cos ^{5} ( 7 x )}{5}) + C

\frac{d y}{d x} = 32 x^{3} y^{3}

x \to 0 lim ((1 - \frac{1}{2 x})^{x})

\frac{d}{d y} (y e^{x} + 2 e^{x} + y^{2})

\int (2 x + 1)^{5} d x

\frac{- 2}{( s ^{2} + 1 ) ^{2}}

What is the integral of sin^5(7x) ?
The integral of sin^5(7x) is 1/7 (-cos(7x)+(2cos^3(7x))/3-(cos^5(7x))/5)+C