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Question Number 130486 by benjo_mathlover last updated on 26/Jan/21

$${\int }_0^x\frac{\cos t\sqrt[4]{{\sin }^{3}t}}{{(\sin x-\sin t)}^{3/4}}dt?$$

Answered by MJS_new last updated on 26/Jan/21

$$\int \cos t{\left(\frac{\sin t}{\sin x-\sin t}\right)}^{3/4}dt=$$$$[u={\left(\frac{\sin t}{\sin x-\sin t}\right)}^{1/4}\to dt=\frac{4{\left({(\sin x-\sin t)}^5{\sin }^{3}t\right)}^{1/4}}{\sin x\cos t}du]$$$$=4\sin x\int \frac{u^6}{{(u^4+1)}^2}du=$$$$\left[\mathrm{Ostrogradski}\right]$$$$=\sin x(-\frac{u^3}{u^4+1}+3\int \frac{u}{u^4+1}du)=$$$$[3\int \frac{u}{u^4+1}du=\frac{3\sqrt{2}}{4}\int (\frac{u}{u^2-\sqrt{2}u+1}-\frac{u}{u^2+\sqrt{2}u+1})du]$$$$=\sin x(-\frac{u^3}{u^4+1}+\frac{3\sqrt{2}}{8}(\ln \frac{u^2-\sqrt{2}u+1}{u^2+\sqrt{2}u+1}+2(\arctan (\sqrt{2}u-1)+\arctan (\sqrt{2}u+1))))$$$$...$$$$t=0\Rightarrow u=0$$$$t=x\Rightarrow u=+\mathrm{\infty }(?)$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{3\pi \sqrt{2}}{4}\sin x(?)$$

Commented by benjo_mathlover last updated on 26/Jan/21

$$amazing$$

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