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प्रश्न
` = ∫ root (3){ cos^2 x} sin x dx `
योग
उत्तर
\[\int \left( \cos^2 x \right)^\frac{1}{3} \sin x dx\]
\[Let, \cos x = t\]
\[ \Rightarrow - \ sin x = \frac{dt}{dx}\]
\[ \Rightarrow \text{sin x dx} = - dt\]
\[Now, \int \left( \cos^2 x \right)^\frac{1}{3} \text{sin x dx}\]
\[ = - \int t^\frac{2}{3} dt\]
\[ = - \left[ \frac{t^\frac{2}{3} + 1}{\frac{2}{3} + 1} \right] + C\]
\[ = - \frac{3}{5} t^\frac{5}{3} + C\]
\[ = - \frac{3}{5} \cos^\frac{5}{3} x + C\]
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