Cho \(I=\int_{0}^{\frac{x}{4}} x \tan ^{2} x d x=\frac{\pi}{a}-\ln \sqrt{b}-\frac{\pi^{2}}{32}\) khi đó tổng a +b bằng
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiTa có: \(I=\int_{0}^{\frac{\pi}{4}} x\left(\frac{1}{\cos ^{2} x}-1\right) d x=\int_{0}^{\frac{\pi}{4}} x \cdot \frac{1}{\cos ^{2} x} d x-\int_{0}^{\frac{\pi}{4}} x d x=I_1+I_2\)
\(I_2=\int_{0}^{\frac{\pi}{4}} x d x=\left.\frac{\pi}{2}\right|_{0} ^{\frac{\pi}{4}}=\frac{\pi^{2}}{32}\)
\(I_{1}=\int_{0}^{\frac{\pi}{4}} x \cdot \frac{1}{\cos ^{2} x} d x\)
Đặt \(\left\{\begin{array}{l} u=x \\ d v=\frac{d x}{\cos ^{2} x} \end{array} \Rightarrow\left\{\begin{array}{l} d u=d x \\ v=\tan x \end{array}\right.\right.\)
Khi đó:
\(I_{1}=\left.x \tan x\right|_{0} ^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \tan x d x=\frac{\pi}{4}+\left.\ln |\cos x|\right|_{0} ^{\frac{\pi}{4}}=\frac{\pi}{4}-\ln \sqrt{2}\)
Vậy \(I=\frac{\pi}{4}-\ln \sqrt{2}-\frac{\pi^{2}}{32}\)
\(\Rightarrow a=4, b=2\Rightarrow a+b=6\)