giúp mình câu 82 nguyên hàm gấp với

Đáp án:

Giải thích các bước giải:

 `f_{1}(x)=sin^2 x=\frac{1-cos\ 2x}{2}`

`F_{1}(x)=\int\ f(x)`

`F_{1}(x)=\int\ \frac{1-cos\ 2x}{2}\ dx`

`F_{1}(x)=\int\ \frac{1}{2}\ dx-\frac{1}{2}\int\ cos\ 2x\ dx`

`F_{1}(x)=1/2 x-1/4sin\ 2x+C`

Mà `F_{1}(0)=0`

`⇔ 1/2 .0-1/4sin\ (2.0)+C=0`

`⇔ C=0`

`⇒ F_{1}(x)=1/2 x-1/4sin\ 2x`

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`f_{2}(x)=cos^2 x=\frac{1+cos\ 2x}{2}`

`F_{2}(x)=\int\ f(x)`

`F_{2}(x)=\int\ \frac{1+cos\ 2x}{2}\ dx`

`F_{2}(x)=\int\ \frac{1}{2}\ dx+\frac{1}{2}\int\ cos\ 2x\ dx`

`F_{2}(x)=1/2 x+1/4sin\ 2x+C`

Mà `F_{2}(0)=0`

`⇔ 1/2 .0+1/4sin\ (2.0)+C=0`

`⇔ C=0`

`⇒ F_{2}(x)=1/2 x+1/4sin\ 2x`

Ta có: `F_{1}(x)=F_{2}(x)`

`⇔ 1/2 x-1/4sin\ 2x=1/2 x+1/4sin\ 2x`

`⇔ 1/2 sin\ 2x=0`

`⇔ sin\ 2x=0`

`⇔ 2x=k\pi\ (k \in \mathbb{Z})`

`⇔ x=k\frac{pi}{2}\ (k \in \mathbb{Z})`