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

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 at 11:27
• wolframalpha.com/input/… – lab bhattacharjee Jan 10 at 11:34
• – lab bhattacharjee Jan 10 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 at 11:42
• @Santos, we can remove $x$ using the link supplied above – lab bhattacharjee Jan 10 at 11:47

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

• what difference it makes, still if I take $t=\sin x$ I will still get $0$,right ? – ss1729 Jan 10 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 at 12:18
• breaking the domain and integrating separately 'd work ? – ss1729 Jan 10 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 at 12:26
• actually I didnt get what you are suggesting ?. $0$ is not the solution right ? – ss1729 Jan 10 at 12:27

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 $$\cos(2\pi-x) = \cos x$$

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

• That is not the function that you were supposed to integrate here. – José Carlos Santos Jan 10 at 11:43
• This is the question, no? – Dhanvi Sreenivasan Jan 10 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 at 11:50
• Yeah, so I simplified it to remove the x term – Dhanvi Sreenivasan Jan 10 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. – ss1729 Jan 10 at 12:09

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?

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