Question

integrate.sin^3(root x)?

Answer 1

Please see the attachment

Answer 2

Use $sin 3t = 3sint - 4sin^3t$

Step-by-step explanation:

$\int sin^3\sqrt x dx$ put $\sqrt x = t$

$\frac {1}{2\sqrt x} = \frac {dt}{dx}$

$dx = 2t dt$

$2\int sin^3t dt$

we know that, $sin 3t = 3sint - 4sin^3t$

$sin^3t = \frac {3sint - sin3t}{4}$

= $2\int t\frac{3sint - sin3t}{4}dt$

= $\frac {3}{2} \int tsintdt - \frac {1}{2}\int tsin3t dt$

= $\frac {3}{2} [t(-cost)-\int(-cost)dt]-\frac {1}{2}[t(\frac {-cos3t}{3})-\int (-\frac {-cos3t}{3})dt]$

= $\frac {3}{2} [-tcost + sint] - \frac {1}{2}[-\frac {-tcos3t}{3}+\frac {1}{3}\frac {sin3t}{3}]+C$

= $\frac {3}{2} [-tcost + sint] + \frac {1}{6}[tcos3t-\frac {sin3t}{3}]+C$

= $\frac {3}{2}\sqrt xcos\sqrt x -\frac {3}{2}sin \sqrt x + \frac {1}{6}[\sqrt xcos3\sqrt x - \frac {sin3\sqrt x}{3}] + C$