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Question Number 219384 by mnjuly1970 last updated on 23/Apr/25

Answered by SdC355 last updated on 24/Apr/25

$$\int \frac{\mathrm{d}x}{\sqrt{x}}{\cos }^3\left(x\right)\sin \left(x\right)=I$$$$\int \frac{\mathrm{d}x}{x}\sqrt{x}{\cos }^3\left(x\right)\sin \left(x\right)=\int \frac{\mathrm{d}x}{x}\sqrt{x}{\cos }^2\left(x\right)\cos \left(x\right)\sin \left(x\right)$$$$\int \frac{\mathrm{d}x}{2x}\sqrt{x}{\cos }^2\left(x\right)\sin \left(2x\right)=\int \mathrm{d}x\frac{e^{-xt}}{x}\sqrt{x}{\cos }^2\left(x\right)\sin \left(2x\right)$$$${\int }_t^{\mathrm{\infty }}\mathrm{d}w{\mathcal{L}}_w\left\{\sqrt{x}{\cos }^2\left(x\right)\sin \left(2x\right)\right\}=$$$$\frac{\sqrt{\pi }}{32}\mathit{i}{\int }_t^{\mathrm{\infty }}\mathrm{d}w\{\frac{2}{\sqrt{{(w+2\mathit{i})}^3}}-\frac{1}{\sqrt{{(w-4\mathit{i})}^3}}+\frac{1}{\sqrt{{(w+4\mathit{i})}^3}}-\frac{2}{\sqrt{{(w-2\mathit{i})}^3}}\}$$$$={[-\frac{\sqrt{\pi }}{32}\mathit{i}\{\frac{4}{\sqrt{w+2\mathit{i}}}-\frac{2}{\sqrt{w-4\mathit{i}}}+\frac{2}{\sqrt{w+4\mathit{i}}}-\frac{4}{\sqrt{w-2\mathit{i}}}\}]}_{w=t}^{w=\mathrm{\infty }}$$$$\frac{\sqrt{\pi }}{32}\mathit{i}\{\frac{4}{\sqrt{t+2\mathit{i}}}-\frac{2}{\sqrt{t-4\mathit{i}}}+\frac{2}{\sqrt{t+4\mathit{i}}}-\frac{4}{\sqrt{t-2\mathit{i}}}\}$$$$t=0$$$$\frac{\sqrt{\pi }}{32}\mathit{i}\{\frac{4}{\sqrt{2\mathit{i}}}-\frac{2}{\sqrt{-4\mathit{i}}}+\frac{2}{\sqrt{4\mathit{i}}}-\frac{4}{\sqrt{-2\mathit{i}}}\}=\frac{4+\sqrt{2}}{32}\sqrt{\pi }$$$$\therefore \alpha =\frac{4+\sqrt{2}}{32}$$

Commented by mnjuly1970 last updated on 23/Apr/25

$$thanksalotbravosir$$

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