The Student Room Group

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Reply 1
Do you have to use substitution?
Reply 2
So I'm told
LIZZIE01
Reply 3
LIZZIE01
Hi,

Sorry that I have not taken the time yet to use the correct text protocol, but can you help me to integrate Sin x cubed, by substitution?

Thanks Lizzie01

Which do you mean A or B

Reply 4
steve I think we can assume it is B as the other one would probably not be A-level, or at least I cannot see a way of doing it... Or well, I guess it is better OP makes clear, but at least I can't possibly spot a substitution for A...
Reply 5
Assuming B, as it has an easy substitution, this should make the substitution obvious:
sin3x dx=(1cos2x)sinx [br]dx\int{\sin^3 x\ \mathrm{d}x} = \int{\left(1 - \cos^2 x\right)\sin x \ [br]\mathrm{d}x}

A, I think, is going to be a more ugly integration by parts problem - unless there's an easier approach that I haven't spotted.
Reply 6
lol. looked up the solution of the version A on maple and it was revolting :smile:
Reply 7
I noticed the same in Mathematica...scary unintelligible answer:p:
Reply 8
Mathmoid
sin3x dx=(1cos2x)sinx [br]dx\int{\sin^3 x\ \mathrm{d}x} = \int{\left(1 - \cos^2 x\right)\sin x \ [br]\mathrm{d}x}

I was going to suggest that, but I wasn't sure where the substitution comes in. The trig identity turns it into a couple of straightforward integrals. :confused:

mathsdave
lol. looked up the solution of the version A on maple and it was revolting :smile:

Yep, it's not nice at all:

sin(x3)dx[br][br]=1/6π23(3/4x22/3sin[br](x3)π9/4x22/3[br](cos(x3)x3sin(x3[br]))π[br][br]3/4x722/3sin(x[br]3)LommelS1(5/6,3/2,x3)1[br]π(x3)116[br][br]+9/4x7[br]22/3(cos(x3)x3sin(x[br]3))LommelS1(116,1/2,x3[br])1π(x3)[br]176)[br]\int\sin(x^3)\,dx[br][br]=1/6\,\sqrt {\pi }\sqrt [3]{2} \left( 3/4\,{\frac {x{2}^{2/3}\sin[br] \left( {x}^{3} \right) }{\sqrt {\pi }}}-9/4\,{\frac {x{2}^{2/3}[br] \left( \cos \left( {x}^{3} \right) {x}^{3}-\sin \left( {x}^{3}[br] \right) \right) }{\sqrt {\pi }}}[br][br]-3/4\,{x}^{7}{2}^{2/3}\sin \left( {x[br]}^{3} \right) {\it LommelS1} \left( 5/6,3/2,{x}^{3} \right) {\frac {1}[br]{\sqrt {\pi }}} \left( {x}^{3} \right) ^{-{\frac {11}{6}}}[br][br]+9/4\,{x}^{7[br]}{2}^{2/3} \left( \cos \left( {x}^{3} \right) {x}^{3}-\sin \left( {x}^[br]{3} \right) \right) {\it LommelS1} \left( {\frac {11}{6}},1/2,{x}^{3}[br] \right) {\frac {1}{\sqrt {\pi }}} \left( {x}^{3} \right) ^{-{\frac {[br]17}{6}}} \right) [br]
Reply 9
LIZZIE01
Hi,

Sorry that I have not taken the time yet to use the correct text protocol, but can you help me to integrate Sin x cubed, by substitution?

Thanks Lizzie01


Here is my attempt. Again it seems rather long. I'm sure someone will be know a better method.
I started a lame attempt, but I seem to be too tired to make anything reasonable out of it...

I do see that the latter integral is 2 I, but am too thick to figure out how to finish this off...
nota bene
I started a lame attempt, but I seem to be too tired to make anything reasonable out of it...


This attempt seems better
Reply 12
I'd really go with Mathmoid's approach:

Unparseable latex formula:

\int{\sin^3 x\ \mathrm{d}x} = \int{\left(1 - \cos^2 x\right)\sin x\,\mathrm{d}x} = \int\sin x\,\mathrm{d}x} - \int{\sin x \cos^2 x\,\mathrm{d}x}

...
steve2005
This attempt seems better

yes, that is easier=)
nota bene
yes, that is easier=)


Yes the second method was much easier. For fun I have done more on my first solution and thankfully I end up with the same result.
Yes, both solutions seems correct=) And I think my attempt with by parts will land there as well when I look at it again tomorrow...

What is strange though is that I ran this in mathematica and it suggested another answer...(haven't checked if they are equivalent):
112(cos(3x)9cos(x))\frac{1}{12}(cos(3x)-9cos(x))

I did differentiate the answer you've got to and it is correct, so we should be safe=)
Reply 16
Just wondering, has anyone ever seen this shorthand for substitution without substitution before?

Unparseable latex formula:

\int{\sin^3x\ \mathrm{d}x} &=& \int{\left(1 - \cos^2x\right)\sin x\ \mathrm{d}x} \\ = \int{\left(\cos^2x - 1\right)\mathrm{d}\left(\cos x\right)} = \frac{1}{3}\cos^3 x - \cos x + c

Reply 17
I haven't, but I like it...
Reply 18
It's "cheating" to say that:

d(cosx)=sinx dx\mathrm{d}\left(\cos x\right) = - \sin x\ \mathrm{d}x

but you can make it rigorous as it is exactly the same process as you go through to do a "real" substitution.
Reply 19
James Gurung
I was going to suggest that, but I wasn't sure where the substitution comes in. The trig identity turns it into a couple of straightforward integrals. :confused:


Yep, it's not nice at all:

sin(x3)dx[br][br]=1/6π23(3/4x22/3sin[br](x3)π9/4x22/3[br](cos(x3)x3sin(x3[br]))π[br][br]3/4x722/3sin(x[br]3)LommelS1(5/6,3/2,x3)1[br]π(x3)116[br][br]+9/4x7[br]22/3(cos(x3)x3sin(x[br]3))LommelS1(116,1/2,x3[br])1π(x3)[br]176)[br]\int\sin(x^3)\,dx[br][br]=1/6\,\sqrt {\pi }\sqrt [3]{2} \left( 3/4\,{\frac {x{2}^{2/3}\sin[br] \left( {x}^{3} \right) }{\sqrt {\pi }}}-9/4\,{\frac {x{2}^{2/3}[br] \left( \cos \left( {x}^{3} \right) {x}^{3}-\sin \left( {x}^{3}[br] \right) \right) }{\sqrt {\pi }}}[br][br]-3/4\,{x}^{7}{2}^{2/3}\sin \left( {x[br]}^{3} \right) {\it LommelS1} \left( 5/6,3/2,{x}^{3} \right) {\frac {1}[br]{\sqrt {\pi }}} \left( {x}^{3} \right) ^{-{\frac {11}{6}}}[br][br]+9/4\,{x}^{7[br]}{2}^{2/3} \left( \cos \left( {x}^{3} \right) {x}^{3}-\sin \left( {x}^[br]{3} \right) \right) {\it LommelS1} \left( {\frac {11}{6}},1/2,{x}^{3}[br] \right) {\frac {1}{\sqrt {\pi }}} \left( {x}^{3} \right) ^{-{\frac {[br]17}{6}}} \right) [br]

OMG, what is LommelS1 etc.?