Tính tích phân sau: A = x dx / cos^2 x

Câu hỏi:

Tính tích phân sau: $A={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{xdx}{{\cos }^{2}x}$

A. $\frac{\mathrm{\pi }}{4}-\ln \frac{\sqrt{2}}{2}$

B. $-\frac{\mathrm{\pi }}{4}+\ln \frac{\sqrt{2}}{2}$

C. $\frac{\mathrm{\pi }}{4}+\ln \frac{\sqrt{2}}{2}$

D. $-\frac{\mathrm{\pi }}{4}-\ln \frac{\sqrt{2}}{2}$

Trả lời:

Chọn C

Đặt $\left\{\begin{array}{l}u=x\\ dv=\frac{dx}{{\cos }^{2}x}\end{array}\Rightarrow \left\{\begin{array}{l}du=dx\\ v=\tan x\end{array}\right.\right.$

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{xdx}{{\cos }^{2}x}=\left(x \tan x\right) {|}_0^{\frac{\mathrm{\pi }}{4}}-{\int }_0^{\frac{\mathrm{\pi }}{4}}\tan xdx \\ =\frac{\mathrm{\pi }}{4}-{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\sin x}{\cos x}dx=\frac{\mathrm{\pi }}{4}+\left(\ln \left|\cos x\right|\right){|}_0^{\frac{\mathrm{\pi }}{4}} \\ =\frac{\mathrm{\pi }}{4}+\left(\ln \frac{\sqrt{2}}{2}-\ln 1\right)=\frac{\mathrm{\pi }}{4}+\ln \frac{\sqrt{2}}{2} \end{gathered}$$