Đáp án+Giải thích các bước giải:

$\displaystyle \int \left(\dfrac{\sin x}{3}+\dfrac{\cos x}{2}\right) \, dx\\ =\displaystyle \int \dfrac{\sin x}{3} \, dx+\displaystyle \int \dfrac{\cos x}{2} \, dx\\ =\dfrac{1}{3}\displaystyle \int \sin x \, dx+\dfrac{1}{2}\displaystyle \int \cos x \, dx\\ =-\dfrac{1}{3}\cos x \, dx+\dfrac{1}{2}\sin x+C\\ \displaystyle \int \dfrac{\sin x-\cos x}{2} \, dx =\dfrac{1}{2}\displaystyle \int \sin x \, dx- \dfrac{1}{2}\displaystyle \int \cos x \, dx =-\dfrac{1}{2}\cos x-\dfrac{1}{2}\sin x+C\\ \displaystyle \int \dfrac{\sin x-4\cos x}{3} \, dx =\dfrac{1}{3}\displaystyle \int \sin x \, dx- \dfrac{4}{3}\displaystyle \int \cos x \, dx =-\dfrac{1}{3}\cos x-\dfrac{4}{3}\sin x+C\\ \displaystyle \int \dfrac{1+\cos^3x}{\cos^2x} \, dx =\displaystyle \int \left(\dfrac{1}{\cos^2x} +\cos x \right)\, dx =\tan x + \sin x +C\\ \displaystyle \int \dfrac{\sin^3x+1}{\sin^2x} \, dx =\displaystyle \int \left(\sin x+ \dfrac{1}{\sin^2x} \right)\, dx =-\cos x-\cot x+C.$