Cho hàm số thỏa mãn f'(x)sinx= f(x)cosx+ 2(sinx)^2

* Hướng dẫn giải

Đáp án D

Tacó:

$f^{\text{'}}\left(x\right)\sin x=f\left(x\right)cos\text{x\hspace{0.17em}}+2{\sin }^{2}x.cos3\text{x\hspace{0.17em}};\forall x\in \left(0;\pi \right)$

$\iff \frac{f^{\text{'}}\left(x\right)\sin x-f\left(x\right)cos\text{x\hspace{0.17em}}}{{\sin }^{2}x}=2cos3\text{x\hspace{0.17em}}$

$\Rightarrow {\left(\frac{f\left(x\right)}{\sin x}\right)}^{\text{'}}=2cos3\text{x}\Rightarrow \frac{f\left(x\right)}{\sin x}=\int 2cos3\text{x\hspace{0.17em}dx}$

$\Rightarrow \frac{f\left(x\right)}{\sin x}=\frac{2}{3}\sin 3\text{x\hspace{0.17em}}+C_1$

$f\left(\frac{\pi }{4}\right)=\frac{1}{3}\Rightarrow C_1=0\Rightarrow f\left(x\right)=\frac{2}{3}sinx.\sin 3x$

$\Rightarrow \int f\left(x\right)\text{d}x=\int \frac{2}{3}sinx.\sin 3x\text{\hspace{0.17em}d}x\text{\hspace{0.17em}=\hspace{0.17em}}\frac{1}{3}\int \left(cos2x-\cos 4x\right)\text{dx\hspace{0.17em}}$

$=\frac{1}{12}(2\sin 2\text{x\hspace{0.17em}\hspace{0.17em}}-\sin 4\text{x) \hspace{0.17em}+\hspace{0.17em}C}$