Question

Biết kết quả tích phân\(I=\)\(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{\cos x}{\sin x+1}dx=aln2+bln3\) với \(a,b\) nguyên.Gía trị của \(H=a.b\) là

Answer 1

Đặt \(sinx=t\Rightarrow cosx.dx=dt\) ; \(\left\{{}\begin{matrix}x=\dfrac{\pi}{6}\Rightarrow t=\dfrac{1}{2}\\x=\dfrac{\pi}{2}\Rightarrow t=1\end{matrix}\right.\)

\(I=\int\limits^1_{\dfrac{1}{2}}\dfrac{dt}{1+t}=ln\left|1+t\right||^1_{\dfrac{1}{2}}=ln2-ln\left(\dfrac{3}{2}\right)=-ln3+2ln2\)

\(\Rightarrow ab=-2\)