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प्रश्न
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
बेरीज
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उत्तर
∫ x . cos3 x2 sin x2 dx
Let x2 = t
⇒ 2x dx = dt
\[\Rightarrow \text{ x dx } = \frac{dt}{2}\]
\[Now, \int x . \cos^3 x^2 \sin x^2 dx\]
\[ = \frac{1}{2}\int \cos^3 t . \sin t . dt\]
\[\text{ Again let }\cos t = p\]
\[ \Rightarrow - \text{ sin t dt } = dp\]
\[ \Rightarrow \text{ sin t dt } = - dp\]
\[So, \frac{1}{2}\int \cos^3 t . \sin t . dt \]
\[ = - \frac{1}{2} p^3 \text{ dp }\]
\[ = - \frac{1}{2} \left( \frac{p^4}{4} \right) + C\]
\[ = - \frac{p^4}{8} + C\]
\[ = - \frac{\cos^4 t}{8} + C\]
\[ = - \frac{\cos^4 x^2}{8} + C\]
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