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integrate :

\int x^{3}sin^{^{4}}xdx

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Let y = \int x^3*sin(x^4) dx

Let us assume that u = x^4

                               du = 4 x^3 dx

                               x^3 dx = (1/4) du

y = (1/4) \int sin(u) du

y = - cos(u) /4 + C                     [ Since\int sin(u) du = - cos(u) + c ]

Therefore y = \int x^3*sin(x^4) dx= - cos(x^4) /4 + C

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Harsh

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