This question already has an answer here:

$$\int \sin^{4}x\cos^{2}xdx$$

$$\int \sin^{4}x\cos^{2}xdx=\int (\sin x \cos x)^{2}\sin^2xdx=\int \left(\frac{\sin^{2}2x}{2}\right)\left(\frac{1}{2}-\frac{\cos2x}{2}\right)dx=\int \left(\frac{\sin^{2}2x}{4}-\frac{\sin^{2}2x\cos2x}{4}\right)dx=\frac{1}{4}\int ({\sin^{2}2x}-{\sin^{2}2x\cos2x})dx$$

I still have not mange to find $u$ substitution


marked as duplicate by Hans Lundmark, user223391, user296602, Shailesh, user228113 Feb 11 '16 at 0:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ How did you obtain the first equality (where did the extra x come from). I recommend you use brackets. $\endgroup$ – George Feb 10 '16 at 20:03
  • 3
    $\begingroup$ uk.answers.yahoo.com/question/index?qid=20090204203206AAbjUfM $\endgroup$ – George Feb 10 '16 at 20:06
  • $\begingroup$ Indeed, there seem to be many typos in the above work, some of which don't seem to affect the flow of the work however and are likely transcription errors. The formatting is also (at least on my screen) wrapping over the edge of the window. I would recommend using multiple lines to display your work instead of trying to write it all on one line. If you wish to be fancy, perhaps use \begin{array}{rl} ...\\ ...\\...\end{array} to improve formatting. $\endgroup$ – JMoravitz Feb 10 '16 at 20:08
  • $\begingroup$ $dx$ is missing $\endgroup$ – Dr. Sonnhard Graubner Feb 10 '16 at 21:00

$$\int\sin^4(x)\cos^2(x)\space\text{d}x=$$ $$\int\sin^4(x)\left(1-\sin^2(x)\right)\space\text{d}x=$$ $$\int\left(\sin^4(x)-\sin^6(x)\right)\space\text{d}x=$$ $$\int\sin^4(x)\space\text{d}x-\int\sin^6(x)\space\text{d}x=$$

You've to use twice the reduction formula:


$$\frac{\sin^5(x)\cos(x)}{6}+\frac{1}{6}\int\sin^4(x)\space\text{d}x=$$ $$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\int\sin^2(x)\space\text{d}x=$$ $$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\int\left[\frac{1}{2}-\frac{\cos(2x)}{2}\right]\space\text{d}x=$$ $$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{1}{2}\int1\space\text{d}x-\frac{1}{2}\int\cos(2x)\space\text{d}x\right]=$$ $$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(2x)\space\text{d}x\right]=$$

Substitute $u=2x$ and $\text{d}u=2\space\text{d}x$:

$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(u)\space\text{d}u\right]=$$ $$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(u)\space\text{d}u\right]=$$ $$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{\sin(u)}{4}\right]+\text{C}=$$ $$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{\sin(2x)}{4}\right]+\text{C}$$


$$\cos^2x = 1 - \sin^2x$$

Known this and the game is done. Then you have to compute

$$\int\sin^4(x) - \sin^6(x)\ \text{d}x$$

Which is quite easy. Can you proceed?


Reduction formula

$$\int\sin^m(x)\ \text{d}x = -\frac{\cos(x) \sin^{n-1}(x)}{n} + \frac{n-1}{n}\int \sin^{n-2}(x)\text{d}x$$

  • $\begingroup$ It is probably because you posted essentially the same hint as Jan (but afterwards) $\endgroup$ – George Feb 10 '16 at 20:16
  • $\begingroup$ @George This is a silly reason. Down votes are for useless / wrong answers, not for correct ones. Also, I'm not that fast in writing with LaTeX and this does explain the amount to time it took me to write the answer.. Anyway fine, who cares. :) $\endgroup$ – Von Neumann Feb 10 '16 at 20:17
  • $\begingroup$ Yes I know, but some people do not understand others can be typing an answer at the same time as someone else. $\endgroup$ – George Feb 10 '16 at 20:18
  • $\begingroup$ @George Exactly! ^^ Also there are some LaTeX commands here which are different from the ones I use in paper documents, and lots of times I have to search for small "errors" xD Anyway.. no problem I'll survive the same lol $\endgroup$ – Von Neumann Feb 10 '16 at 20:21
  • $\begingroup$ "Known this and the game is done" thanks $\endgroup$ – gbox Feb 10 '16 at 20:31

Your own approach was easier than some of those suggested. Picking up from where you left off, you just use the double angle formula and reversing the chain rule, and we have $$\frac 14\int \frac 12(1-\cos 4x)dx-\frac 14\times \frac 16\sin^32x$$ Can you finish this now?


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