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प्रश्न
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उत्तर
\[\text{ Let I} = \int x \cdot \sin^5 x^2 \cdot \cos x^2 \text{ dx }\]
\[\text{ Putting sin x}^2 = t\]
\[ \Rightarrow \text{ cos} \left( x^2 \right) \times 2x \text{ dx } = dt\]
\[ \Rightarrow \text{ cos} \left( x^2 \right) \cdot x \text{ dx }= \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int t^5 \cdot dt\]
\[ = \frac{1}{2} \left[ \frac{t^6}{6} \right] + C\]
\[ = \frac{t^6}{12} + C\]
\[ = \frac{\sin^6 x^2}{12} + C .....\left[ \because t = \sin x^2 \right]\]
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