Biết

A

- Hướng dẫn giải

Đáp án A.

Đặt $\left\{\begin{array}{l}u=x\\ dv=\frac{dx}{{\sin }^{2}x}\end{array}\iff \right.\left\{\begin{array}{l}du=dv\\ v=-cotx\end{array}\right.$, khi đó ${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{x}{{\sin }^{2}x}dx={\overline{)\left(-x.cotx\right)}}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}+{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}cotxdx$ .

Xét tích phân  ${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}cotxdx={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{\cos x}{\sin x}dx={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{d\left(\sin x\right)}{\sin x}={\overline{)\ln \left(\sin x\right)}}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}=-\ln \frac{\sqrt{2}}{2}.$ 

Vậy $I={\overline{)\left(-x.cotx\right)}}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}-\ln \frac{\sqrt{2}}{2}=\frac{\mathrm{\pi }}{4}-\ln \frac{\sqrt{2}}{2}=\frac{1}{4}\mathrm{\pi }+\frac{1}{2}.\ln 2=\mathrm{m}.\mathrm{\pi }+\mathrm{n}.\ln 2\Rightarrow \left\{\begin{array}{l}\mathrm{m}=\frac{1}{4}\\ \mathrm{n}=\frac{1}{2}\end{array}\Rightarrow \mathrm{P}=1.\right.$