Tìm II= nguyên hàm cos^4 x/ sin ^4 x + cos ^4 x dx ?

Tìm $I=\int \frac{{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$?

Hướng dẫn:

Đặt : $T=\int \frac{{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$\Rightarrow I+T=\int \frac{{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx+\int \frac{{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx=\int \frac{{\sin }^{4}x+{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx=x+C_1\left(1\right)$

Mặt khác :

$\begin{array}{l}I-T=\int \frac{{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx-\int \frac{{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx=\int \frac{{\cos }^{4}x-{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx\\ \iff I-T=\int \frac{{\cos }^{2}x-{\sin }^{2}x}{1-2{\sin }^{2}x.{\cos }^{2}x}dx=\int \frac{\cos 2x}{1-\frac{1}{2}{\sin }^{2}x}dx\\ \iff I-T=\int \frac{2\cos 2x}{2-{\sin }^{2}2x}dx=\frac{1}{2\sqrt{2}}\ln \left(\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right)+C_2\left(2\right)\end{array}$

Từ $\left(1\right);\left(2\right)$ ta có hệ : $\left\{\begin{array}{l}I+T=x+C_1\\ I-T=\frac{1}{2\sqrt{2}}\ln \left(\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right)+C_2\end{array}\right.\Rightarrow \left\{\begin{array}{l}I=\frac{1}{2}\left(x+\frac{1}{2\sqrt{2}}\ln \left(\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right)\right)+C\\ T=\frac{1}{2}\left(x-\frac{1}{2\sqrt{2}}\ln \left(\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right)\right)+C\end{array}\right.$

Vậy đáp án đúng là đáp án C.