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