Đáp án:
$I=\dfrac{1}{2}\cos\dfrac{x}{2}$
Giải thích các bước giải:
$I=\dfrac{2\sin x-1+2\cos^2\left (\dfrac{\pi}{4}+\dfrac{x}{2} \right )}{4\sin\dfrac{x}{2}}\\
=\dfrac{2\sin x+\cos\left (\dfrac{\pi}{2}+x \right )}{4\sin\dfrac{x}{2}}\\
=\dfrac{2\sin x-\sin x}{4\sin\dfrac{x}{2}}\\
=\dfrac{\sin x}{4\sin\dfrac{x}{2}}\\
=\dfrac{2\sin \dfrac{x}{2}\cos\dfrac{x}{2}}{4\sin\dfrac{x}{2}}\\
=\dfrac{1}{2}\cos\dfrac{x}{2}$
