Question

To evaluate ∫ 3x2 cos (2x3 - 4) dx, it is necessary to let

Answer 1

Try using integration by parts.

Answer 2

If we were to take
`y=2x^3-4`, then we have
`\mathrm dy=6x^2\,\mathrm dx`. So in the integral, we could write


`\displaystyle\int3x^2\cos(2x^3-4)\,\mathrm dx=\int2(3x^2)\cos(2x^3-4)\,\mathrm dx=2\int\cos y\,\mathrm dy=2\sin y+C`

Then back-substituting to get the antiderivative in terms of
`x`, we have


`\displaystyle\int3x^2\cos(2x^3-4)\,\mathrm dx=2\sin(2x^3-4)+C`