Cho hai tích phân \(\int_{0}^{\frac{\pi }{2}}sin^2xdx\) và \(\int_{0}^{\frac{\pi }{2}}cos^2x dx\). Hãy chỉ ra khẳng định đúng:
(A) \(\int_{0}^{\frac{\pi }{2}}sin^2xdx >\int_{0}^{\frac{\pi }{2}}cos^2x dx\)
(B) \(\int_{0}^{\frac{\pi }{2}}sin^2xdx <\int_{0}^{\frac{\pi }{2}}cos^2x dx\)
(C) \(\int_{0}^{\frac{\pi }{2}}sin^2xdx =\int_{0}^{\frac{\pi }{2}}cos^2x dx\)
(D) Không so sánh được
Ta có:
\(\int_{0}^{\frac{\pi }{2}}sin^2xdx=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}(1-cos2x)dx\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}dx-\frac{1}{4}\int_{0}^{\frac{\pi }{2}}cos2x d2x\)
\(=\frac{1}{2} x \bigg|_{0}^{\frac{\pi }{2}}-\frac{1}{4}sin2x\bigg|_{0}^{\frac{\pi }{2}}= \frac{\pi }{4}-0=\frac{\pi }{4}\)
\(\int_{0}^{\frac{\pi }{2}}cos^2 x dx=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}(1+cos2x)dx\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}dx+\frac{1}{4}\int_{0}^{\frac{\pi }{2}} cos2x d2x\)
\(=\frac{1}{2}x \Bigg|^{\frac{\pi }{2}}_0+\frac{1}{4}sin2x\Bigg|^{\frac{\pi }{2}}_0= \frac{\pi }{4}+0=\frac{\pi }{4}\)
\(\Rightarrow \int_{0}^{\frac{\pi }{2}}sin^2x dx= \int_{0}^{\frac{\pi }{2}}cos^2 x dx.\)
Nên khẳng định đúng là (C).
-- Mod Toán 12
Copyright © 2021 HOCTAP247