Can someone please show me with steps on how to evaluate this indefinite integral? enter image description here


Let $u = \sin x \implies du = \cos x\,dx$.

$$\int 3(\,\underbrace{\sin(x)}_{u}\,)^3 \,\underbrace{\cos x\,dx}_{du}= \int 3u^3 \,du$$

  • $\begingroup$ so the integral of 3u^3 would be 3/4u^4 ..? $\endgroup$ – Sherry Eskandarian Mar 1 '14 at 0:13
  • $\begingroup$ Yes, $\frac 34 u^4 + C$. And then back substitute: that gives us $\frac 34 \sin^4 x + C$ $\endgroup$ – Namaste Mar 1 '14 at 0:35

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