CÂU HỎI: Hàm số \(f(x)=\frac{7 \cos x-4 \sin x}{\cos x+\sin x}\)có một nguyên hàm F(x) thỏa mãn \(F\left(\frac{\pi}{4}\right)=\frac{3 \pi}{8}\). Giá trị của \(F\left(\frac{\pi}{2}\right)\) bằng:

\(\begin{array}{l}
\text { Ta có } f(x)=\frac{\frac{3}{2}(\sin x+\cos x)+\frac{11}{2}(-\sin x+\cos x)}{\cos x+\sin x}=\frac{3}{2}+\frac{11}{2} \cdot \frac{-\sin x+\cos x}{\cos x+\sin x} \\
\Rightarrow F(x)=\int f(x) \mathrm{d} x=\int\left(\frac{3}{2}+\frac{11}{2} \cdot \frac{-\sin x+\cos x}{\cos x+\sin x}\right) \mathrm{d} x=\frac{3}{2} x+\int \frac{11}{2} \cdot \frac{-\sin x+\cos x}{\cos x+\sin x} \mathrm{d} x \\
=\frac{3}{2} x+\frac{11}{2} \int \frac{1}{\cos x+\sin x} \mathrm{d}(\cos x+\sin x)=\frac{3}{2} x+\frac{11}{2} \ln |\cos x+\sin x|+C
\end{array}\)

\(\begin{array}{l}
\text { Mà } F\left(\frac{\pi}{4}\right)=\frac{3 \pi}{8} \Rightarrow \frac{3 \pi}{8}+\frac{11}{2} \ln \sqrt{2}+C=\frac{3 \pi}{8} \Rightarrow C=-\frac{11}{4} \ln 2 \\
\text { Do đó } F\left(\frac{\pi}{2}\right)=\frac{3 \pi}{4}+C=\frac{3 \pi}{4}-\frac{11}{4} \ln 2
\end{array}\)