Evaluate the integral. $\int sin^3 (5x) dx$

Question

The bounds are from $(\pi/5)$ to $0$. I know we use a pythagorean identity for this. So $u=\sin^2 (x)$ and $du = \sin (2x) dx$. But I'm helping trouble solving this problem.

Answer 1

$$\int\sin^{2n+1}mx\ dx=\int(1-\cos^2mx)^n\sin mx\ dx$$

Set $\cos mx=u$

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Alternatively, $\displaystyle\sin3x=3\sin x-4\sin^3x\iff\sin^3x=\frac{3\sin x-\sin3x}4$

Answer 2

Hint : $\sin(3x) = 3\sin x - 4\sin^3x $

$\implies \sin (15x) = 3 \sin (5x) - 4 \sin^3( 5x)$

$\implies \sin^3( 5x) = \dfrac {1}{4} [ 3 \sin (5x) - \sin (15x) ]$

$\implies \int \sin^3( 5x) dx = \dfrac {1}{4} [ 3 \int\sin (5x)dx - \int \sin (15x) dx]$

Answer 3

\begin{align*} \int \sin^3(5x) \, dx &= \int \sin^2(5x) \cdot \sin(5x) \, dx \\ &= \int \left( 1-\cos^2(5x) \right)\sin(5x) \, dx \\ &= \int \left( \sin(5x)-\cos^2(5x)\sin(5x) \right) \, dx \\ &= -\frac{\cos(5x)}{5}-\int \cos^2(5x)\sin(5x) \, dx \\ \end{align*}

For the second integral, let $u=\cos(5x)$.

\begin{align*} u &=\cos(5x) \\ \Rightarrow du &=-5\sin(5x) \, dx \\ \Rightarrow -\frac{1}{5} du &= \sin(5x) \, dx. \end{align*}

Hence,

\begin{align*} -\int \cos^2(5x)\sin(5x) \, dx &=\frac{1}{5}\int u^2 \, du \\ &=\frac{u^3}{15}+c \\ &=\frac{\cos^3(5x)}{15}+c. \end{align*}

Putting it all together,

$$\int \sin^3(5x) \, dx =-\frac{\cos(5x)}{5}+\frac{\cos^3(5x)}{15}+c.$$

I leave you to evaluate your limits.