Question

`  =  ∫ root (3){ cos^2 x}  sin x   dx `

Answer 1

\[\int \left( \cos^2 x \right)^\frac{1}{3} \sin x dx\]

\[Let, \cos x = t\]

\[ \Rightarrow - \ sin x = \frac{dt}{dx}\]

\[ \Rightarrow \text{sin x dx} = - dt\]

\[Now, \int \left( \cos^2 x \right)^\frac{1}{3} \text{sin x dx}\]

\[ = - \int t^\frac{2}{3} dt\]

\[ = - \left[ \frac{t^\frac{2}{3} + 1}{\frac{2}{3} + 1} \right] + C\]

\[ = - \frac{3}{5} t^\frac{5}{3} + C\]

\[ = - \frac{3}{5} \cos^\frac{5}{3} x + C\]