(Sở Hà Tĩnh 2022) Cho \(F\left( x \right)\) là nguyên hàm của \(f\left( x \right) = {\sin ^2}x\) trên \(\mathbb{R}\) thoả mãn \(F\left( {\frac{\pi }{4}} \right) = 0\). Giá trị biểu thức \(S = F\left( { – \pi } \right) + 2F\left( {\frac{\pi }{2}} \right)\) bằng

A. \(S = \frac{3}{4} – \frac{\pi }{4}\).

B. \(S = \frac{3}{4} – \frac{{3\pi }}{4}\).

C. \(S = \frac{1}{4} + \frac{{3\pi }}{8}\).

D. \(S = \frac{3}{2} – \frac{{3\pi }}{8}\).

Lời giải:

Chọn B

Ta có \(\int {{{\sin }^2}x} {\rm{d}}x = \int {\frac{{1 – \cos 2x}}{2}} {\rm{d}}x = \frac{1}{2}x – \frac{1}{4}\sin 2x + C\).

\( \Rightarrow F\left( x \right) = \frac{1}{2}x – \frac{1}{4}\sin 2x + C\)

Mà \(F\left( {\frac{\pi }{4}} \right) = 0 \Leftrightarrow \frac{\pi }{8} – \frac{1}{4}\sin \frac{\pi }{2} + C = 0\)\( \Leftrightarrow C = – \frac{\pi }{8} + \frac{1}{4}\).

Vậy \(F\left( x \right) = \frac{1}{2}x – \frac{1}{4}\sin 2x – \frac{\pi }{8} + \frac{1}{4}\).

\(S = F\left( { – \pi } \right) + 2F\left( {\frac{\pi }{2}} \right)\)\( = – \frac{\pi }{2} – \frac{1}{4}\sin \left( { – 2\pi } \right) – \frac{\pi }{8} + \frac{1}{4} + 2\left[ {\frac{1}{2}.\frac{\pi }{2} – \frac{1}{4}\sin 2.\frac{\pi }{2} – \frac{\pi }{8} + \frac{1}{4}} \right]\)\( = – \frac{\pi }{2} – \frac{\pi }{8} + \frac{\pi }{4} + \frac{1}{4} + \frac{1}{2} = \frac{3}{4} – \frac{{3\pi }}{8}\).