nguyên hàm =cos ^2(3x)dx

Đáp án:

$\dfrac{1}{2}\left(x+\dfrac{1}{6}\sin  6x\right)+C.$

Giải thích các bước giải:

$\displaystyle\int\cos^23x \, dx\\ =\dfrac{1}{2}\displaystyle\int (1+\cos 6x) \, dx\\ =\dfrac{1}{2}\left(\displaystyle\int  \, dx+\displaystyle\int \cos 6x \, dx\right)\\ =\dfrac{1}{2}\left(x+\dfrac{1}{6}\displaystyle\int \cos 6x \, d(6x)\right)\\ =\dfrac{1}{2}\left(x+\dfrac{1}{6}\sin  6x\right)+C.$