Ta có
$\dfrac{1}{2}[\sin^2x + \cos^2x + \sin(2x)] \ln(\sin x + \cos x) - \dfrac{1 + \sin(2x)}{4} + C_1$
$= \dfrac{1}{2}[(\sin^2x + \cos^2x) + \sin(2x)] \ln\sqrt{(\sin x + \cos x)^2} - \left( \dfrac{1}{4} + \dfrac{\sin(2x)}{4} \right) + C_1$
$= \dfrac{1}{2}[1 + \sin(2x)] \ln\left[ (\sin x + \cos x)^2 \right]^{\frac{1}{2}} - \dfrac{1}{4} - \dfrac{\sin(2x)}{4} + C_1$
$= \dfrac{1}{2}[1 + \sin(2x)] . \dfrac{1}{2} \ln (\sin^2x + \cos^2x + 2\sin x \cos x) - \dfrac{\sin(2x)}{4} - \dfrac{1}{4} + C_1$
$= \dfrac{1}{4} [1 + \sin(2x)] \ln[(\sin^2x + \cos^2x) + \sin(2x)] - \dfrac{\sin(2x)}{4} - \dfrac{1}{4} + C_1$
$= \dfrac{1}{4} [1 + \sin(2x)] \ln[1 + \sin(2x)] - \dfrac{\sin(2x)}{4} - \dfrac{1}{4} + C_1$
