I was asked to find $$\int \sin^5x \, \mathrm dx$$ only in form of $\cos x$. I solved it using recursion, but I was getting something like $\cos^5x+ \cos^3x+\cos x$(obviously with some coefficients), but the options also had $\cos^4x+\cos^2x$ but I am not getting any, am I making some error, or I need to manipulate my answer. Thanks.
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Show us the recursions you used. – GFauxPas Mar 24 '15 at 16:52
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3What answer did you get, and what is the answer you think you should be getting? – Mark Bennet Mar 24 '15 at 16:52
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$$\newcommand{\co}{\color{grey}{\rm constant}}\newcommand{\d}{{\rm d}}\int\sin^5x\d x\stackrel{t=\cos x}=-\int (1-t^2)^2\d t=\int(1+t^4-2t^2)\d t=-t-t^5/5+2t^3/3+\co\\=-\cos x-(1/5)\cos^5x+(2/3)\cos^3x+\co\\=\frac1{15} \cos(x) (3 \cos^4(x)-10 \cos^2(x)+15)+\co$$
RE60K
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@mathgeek isn't the last form with even powers? check the answer by differentiating or upload it's image/mention exact answer for us – RE60K Mar 24 '15 at 17:23
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Sorry, what second thing? I want my final answer like in your second step. – mathgeek Mar 24 '15 at 17:27
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$$\int\sin^{2n+1}x\ dx=\int(1-\cos^2x)^n\sin x\ dx$$
Set $\cos x=u$
lab bhattacharjee
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@mathgeek, Have you validated the answer, may be by differentiating? This helps you to avoid recursion – lab bhattacharjee Mar 24 '15 at 16:56
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Yes, your answer as well as mine as well as that of WolframAlpha are same and correct, but I think we need some manipulation to get to the required form. Maybe our constant term might become diiferent or something else... – mathgeek Mar 24 '15 at 17:01