Question

\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]

Answer 1

∫ x . cos³ x² sin x² dx
Let x² = t
⇒​ 2x dx = dt

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