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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 ?

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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 $$I=2\pi\int_0^{\pi/2}\sin^6x\cos^5x-\sin^6x\cos^5xdx$$ $$=0$$

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  • $\begingroup$ what difference it makes, still if I take $t=\sin x$ I will still get $0$,right ? $\endgroup$ – ss1729 Jan 10 at 12:12
  • $\begingroup$ @ss1729, You cannot substitute $t = \sin x$ in the original integral because $\sin x$ is not a one-one function. $\endgroup$ – Martund Jan 10 at 12:18
  • $\begingroup$ breaking the domain and integrating separately 'd work ? $\endgroup$ – ss1729 Jan 10 at 12:22
  • $\begingroup$ @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. $\endgroup$ – Martund Jan 10 at 12:26
  • $\begingroup$ actually I didnt get what you are suggesting ?. $0$ is not the solution right ? $\endgroup$ – ss1729 Jan 10 at 12:27
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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 $\cos(2\pi-x) = \cos x$

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

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  • $\begingroup$ That is not the function that you were supposed to integrate here. $\endgroup$ – José Carlos Santos Jan 10 at 11:43
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    $\begingroup$ This is the question, no? $\endgroup$ – Dhanvi Sreenivasan Jan 10 at 11:47
  • $\begingroup$ No. The question is about $\int_0^{2\pi}x\sin^6(x)\cos^5(x)\,\mathrm dx$. $\endgroup$ – José Carlos Santos Jan 10 at 11:50
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    $\begingroup$ Yeah, so I simplified it to remove the x term $\endgroup$ – Dhanvi Sreenivasan Jan 10 at 11:51
  • $\begingroup$ 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. $\endgroup$ – ss1729 Jan 10 at 12:09
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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?

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  • $\begingroup$ 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 $0$ ? $\endgroup$ – ss1729 Jan 10 at 12:56
  • $\begingroup$ Yes, the answer is $0$. $\endgroup$ – José Carlos Santos Jan 10 at 12:58
  • $\begingroup$ actually my reference gives the solution $\dfrac{32\pi}{693}$, so it must be wrong ? $\endgroup$ – ss1729 Jan 10 at 12:59
  • $\begingroup$ Yes, it is wrong. $\endgroup$ – José Carlos Santos Jan 10 at 13:07

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