Giá trị của tích phân \(I=\int_{0}^{\pi} \frac{x d x}{\sin x+1} 1\) là

* Hướng dẫn giải

Đặt \(x=\pi-t \Rightarrow d x=-d t \text { Đổi cận: } x=0 \Rightarrow t=\pi, x=\pi \Rightarrow t=0\)

\(\begin{array}{l} \Rightarrow I=-\int\limits_{\pi}^{0} \frac{(\pi-t) d t}{\sin (\pi-t)+1}=\int\limits_{0}^{\pi}\left(\frac{\pi}{\sin t+1}-\frac{t}{\sin t+1}\right) d t=\pi \int\limits_{0}^{\pi} \frac{d t}{\sin t+1}-I \Rightarrow I=\frac{\pi}{2} \int\limits_{0}^{\pi} \frac{d t}{\sin t+1} \\ =\frac{\pi}{2} \int\limits_{0}^{\pi} \frac{d t}{\left(\sin \frac{t}{2}+\cos \frac{t}{2}\right)^{2}}=\frac{\pi}{4} \int\limits_{0}^{\pi} \frac{d t}{\cos ^{2}\left(\frac{t}{2}-\frac{\pi}{4}\right)}=\frac{\pi}{2} \int\limits_{0}^{\pi} \frac{d\left(\frac{t}{2}-\frac{\pi}{4}\right)}{\cos ^{2}\left(\frac{t}{2}-\frac{\pi}{4}\right)}=\left.\frac{\pi}{2} \tan \left(\frac{t}{2}-\frac{\pi}{4}\right)\right|_{0} ^{\pi}=\pi \end{array}\)