`int3cos^3x\ dx`

`=3int(cos^2x)cos\ x\ dx`

`=3int(1-sin^2x)cos\ x\ dx`

`=3int(cos\ x\ -sin^2x\ cos\ x\) dx`

Letting `u=sin\ x`, and thus `du=cos\ x\ dx` gives:

`=3[sin\ x+K_1-intu^2du]`

`=3[sin\ x-(u^3)/3+K]`

`=3[sin\ x-(sin^3x)/3+K]`

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