Question

solve it please.............​

Answer 1

refer to pic above please

Answer 2

Answer:

$∫sin3xcos2xdx∫sin3xcos2xdx</p><p></p><p>=∫sinx(sin2x)cos2xdx=∫sinx(sin2x)cos2xdx</p><p></p><p>=∫sinx(1−cos2x)cos2xdx=∫sinx(1−cos2x)cos2xdx</p><p></p><p>=∫(sinxcos2x−sinxcos4x)dx=∫(sinxcos2x−sinxcos4x)dx</p><p></p><p>=−∫cos2x(−sinx)dx+∫cos4x(−sinx)dx=−∫cos2x(−sinx)dx+∫cos4x(−sinx)dx</p><p></p><p>=−cos3x3+cos5x5+C=−cos3x3+cos5x5+C</p><p></p><p>=115cos3x(3cos2−5)+C</p><p></p><p>$