Biết tích phân sin4x/(sin^4 x+cos^4 x)=F(x)+c. Khi đó

* Hướng dẫn giải

Đáp án B.

$$\begin{gathered} \frac{\sin 4x}{{\sin }^{4}x+{\cos }^{4}x}=\frac{2\sin 2x\cos 2x}{1-2{\sin }^{2}x{\cos }^{2}x}=\frac{2\sin 2x\cos 2x}{1-2.{\displaystyle \frac{1}{4}}{\sin }^{2}2x}=\frac{4\sin 2x\cos 2x}{2-{\sin }^{2}2x} \\ \Rightarrow \int \frac{\sin 4xdx}{{\sin }^{4}x+{\cos }^{4}x}=\int \frac{4\sin 2x\cos 2xdx}{2-{\sin }^{2}2x}=I \\ Đặt 2-{\sin }^{2}2x=t \iff -4\sin 2x\cos 2xdx=dt \\ \Rightarrow I=\int \frac{-dt}{t}=-\ln \left|t\right|=-\ln \left| 2-{\sin }^{2}2x\right|=\ln \left|\frac{1}{ 2-{\sin }^{2}2x}\right| \\ =\ln \left|\frac{1}{{\sin }^{4}x+{\cos }^{4}x}\right|=\ln \left(\frac{1}{{\sin }^{4}x+{\cos }^{4}x}\right) \left(do \frac{1}{{\sin }^{4}x+{\cos }^{4}x}>0\right) \end{gathered}$$