Evaluate $\int_0^{2\pi}x\sin^6x.\cos^5x.dx$

Question

Evaluate $$ \int_0^{2\pi}x\sin^6x.\cos^5x.dx $$

Set $t=\sin x\implies dt=\cos x.dx$ $$ I=\int_0^{2\pi}(2\pi-x)\sin^6x.\cos^4x.\cos x.dx\\ 2I=2\pi\int_0^{2\pi}\sin^6x.(1-\sin^2x)^2.\cos x.dx\\ I=\pi\int_0^{2\pi}\sin^6x.(1-2\sin^2x+\sin^4x).\cos x.dx=\pi\int_0^{\color{red}{?}}[t^6-2t^8+t^{10}]dt $$ The solution given in my reference is $\dfrac{32\pi}{693}$ but if I set $x:0\to2\pi\implies t:0\to 0$, integral becomes zero, so what exactly should be the uper limit of the definite integral ?

Thanx @José Carlos Santos, $$ I=\pi\bigg[\int_0^{\pi/2}f(x)dx+\int_{\pi/2}^{3\pi/2}f(x)dx+\int_{\pi/2}^{2\pi}f(x)dx\bigg]\\ =\pi\Big[\int_0^{1}\big[\frac{t^7}{7}-\frac{2t^9}{9}+\frac{t^{11}}{11}\big]dt+\int_{1}^{-1}\big[\frac{t^7}{7}-\frac{2t^9}{9}+\frac{t^{11}}{11}\big]dt+\int_{-1}^{0}\big[\frac{t^7}{7}-\frac{2t^9}{9}+\frac{t^{11}}{11}\big]dt\Big]=0 $$ Even If I split the limits, seems like it still gives me $0$, how do I deal with it properly ?

Let $I(t)=\int_0^t f(x) dx$ with $f(x)$ to be the integrand. Then $I(t)-\frac{32 \pi}{693}$ is negative on $[0,2\pi]$! — Math-fun, Jan 10 '20 at 11:27
https://www.wolframalpha.com/input/?i=%5Cint_0%5E%7B%5Cpi%7D%28%5Csin%5E6x%5Ccdot%5Ccos%5E5x%29+dx — lab bhattacharjee, Jan 10 '20 at 11:34
Start with https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 — lab bhattacharjee, Jan 10 '20 at 11:35
@labbhattacharjee The function that is being integrated here is $x\sin^6(x)\cos^5(x)$, not $\sin^6(x)\cos^5(x)$. — José Carlos Santos, Jan 10 '20 at 11:42
@labbhattacharjee my actual doubt is regarding the limit after $x$ is removed. — Sooraj S, Jan 10 '20 at 12:14
@labbhattacharjee my reference gives the solution $\dfrac{32\pi}{693}$, can you confirm it as zero, ie. my reference wrong ? — Sooraj S, Jan 10 '20 at 13:00

Martund · Accepted Answer · 2020-01-10T12:35:41.750

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Note that your substitution is not working because the substitution rule works only for one-one functions, whereas $\sin x$ is not a one-one function in $(0,2\pi)$. Instead, use the fact that $$\int_0^{2a}f(x)dx=\int_0^af(x)+f(2a-x)dx$$ $$\implies I=\int_0^{2\pi}x\sin^6x\cos^5xdx$$ $$=\int_0^\pi(x+2\pi-x)\sin^6x\cos^5xdx$$ $$=2\pi\int_0^\pi\sin^6x\cos^5xdx$$ Again apply the same property and get $$I=2\pi\int_0^{\pi/2}\sin^6x\cos^5x-\sin^6x\cos^5xdx$$ $$=0$$

edited Jan 10 '20 at 12:35

answered Jan 10 '20 at 11:47

Martund

14,706
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what difference it makes, still if I take $t=\sin x$ I will still get $0$,right ? – Sooraj S Jan 10 '20 at 12:12
@ss1729, You cannot substitute $t = \sin x$ in the original integral because $\sin x$ is not a one-one function. – Martund Jan 10 '20 at 12:18
breaking the domain and integrating separately 'd work ? – Sooraj S Jan 10 '20 at 12:22
@ss1729, Yeah, that would work. But the method I am suggesting in my answer is almost mental arithmetic (if you practice this rule), so that would be a lot easier, because you'll need to do integration by parts in each of the term separately. – Martund Jan 10 '20 at 12:26
actually I didnt get what you are suggesting ?. $0$ is not the solution right ? – Sooraj S Jan 10 '20 at 12:27
@ss1729, please tell me specifically which step you didn't understand in my solution. The answer is zero, as you can check on wolfram alpha. – Martund Jan 10 '20 at 12:33
@ss1729, I have edited the answer to include the last step also. – Martund Jan 10 '20 at 12:36
okay, so it seems my reference must be wrong ! – Sooraj S Jan 10 '20 at 13:04

score 1 · Answer 2 · answered Jan 10 '20 at 11:42

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Hint

Applying the rule

$$\int_a^bf(x)dx = \int_a^bf(a+b-x)dx$$

$$I = \int_0^{2\pi}x\sin^6x\cos^5xdx = \int_0^{2\pi}(2\pi-x)\sin^6(2\pi-x)\cos^5(2\pi-x)dx$$

Using the fact that $\sin(2\pi-x) = -\sin x$ and $\cos(2\pi-x) = \cos x$

$$I = \pi\int_0^{2\pi}\sin^6x \cos^5x dx$$

answered Jan 10 '20 at 11:42

Dhanvi Sreenivasan

5,637

That is not the function that you were supposed to integrate here. – José Carlos Santos Jan 10 '20 at 11:43
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This is the question, no? – Dhanvi Sreenivasan Jan 10 '20 at 11:47
No. The question is about $\int_0^{2\pi}x\sin^6(x)\cos^5(x),\mathrm dx$. – José Carlos Santos Jan 10 '20 at 11:50
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Yeah, so I simplified it to remove the x term – Dhanvi Sreenivasan Jan 10 '20 at 11:51
this is something I forgot to put when I typed the question here. The real problem I am confused is about the limit. ie, after your last step. – Sooraj S Jan 10 '20 at 12:09

score 1 · Answer 3 · answered Jan 10 '20 at 11:57

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In order to compute$$\int_0^{\pi/2}x\sin^x(x)\cos^5(x)\,\mathrm dx\tag1$$using the substitution that you mentioned, then, since, when $x$ goes from $0$ to $\frac\pi2$, $\sin(x)$ goes from $0$ to $1$, you get$$\int_0^1\arcsin(t)(t^6-2t^8+t^{10})\,\mathrm dt=\frac{4(-5\,156+3\,465\pi)}{2\,401\,245}.$$So, $(1)=\frac{4(-5\,156+3\,465\pi)}{2\,401\,245}$.

Now, using the same idea in order to compute$$\int_{\pi/2}^{3\pi/2}x\sin^x(x)\cos^5(x)\,\mathrm dx,$$what you get is$$\int_1^{-1}\bigl(\pi-\arcsin(t)\bigr)(t^6-2t^8+t^{10})\,\mathrm dt=-\frac{16\pi}{693}.$$Can you take it from here?

answered Jan 10 '20 at 11:57

José Carlos Santos

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Please check I have modified OP to include attempt to solve by slitting the limits as you showed, but I think it is still giving $0$ ? – Sooraj S Jan 10 '20 at 12:56
Yes, the answer is $0$. – José Carlos Santos Jan 10 '20 at 12:58
actually my reference gives the solution $\dfrac{32\pi}{693}$, so it must be wrong ? – Sooraj S Jan 10 '20 at 12:59
Yes, it is wrong. – José Carlos Santos Jan 10 '20 at 13:07

Evaluate $\int_0^{2\pi}x\sin^6x.\cos^5x.dx$

3 Answers3