Question

int {thita sin^2(thita) cos(thita) dthita}

Answer 1

Step-by-step explanation:

We have,

$\int \{ \theta. \sin^{2} \theta. { \cos }^{2} \theta \}d \theta$

$= \int \{ \frac{ \theta}{4}.(4\sin^{2} \theta{ \cos }^{2} \theta )\}d \theta$

$= \frac{1}{4} \int \theta.\sin^{2} 2\theta \: d \theta$

Using IBP,

$= \frac{1}{4} \sin^{2} 2\theta \int \theta \: d \theta - \frac{1}{4} \int \{ \frac{d}{d\theta }( \sin ^{2}2 \theta) \int \theta \: d \theta \} d \theta$

$= \frac{1}{4} \sin^{2} 2\theta - \frac{1}{4} \int \{4 \sin2 \theta \: \cos2 \theta \} d \theta$

$= \frac{1}{4} \sin^{2} 2\theta - \frac{1}{2} \int \{2 \sin2 \theta \: \cos2 \theta \} d \theta$

$= \frac{1}{4} \sin^{2} 2\theta - \frac{1}{2} \int \sin4 \theta d \theta$

$= \frac{1}{4} \sin^{2} 2\theta - \frac{1}{2} ( \frac{ - \cos4 \theta }{4} )$

$= \frac{1}{4}( \sin^{2} 2\theta + \frac{1}{2} \cos4 \theta ) + c$