Câu hỏi
17/12/2024 1Tính các tích phân sau:
a) \(\int\limits_0^{\frac{\pi }{2}} {\left( {3\cos x + 2\sin x} \right)dx} \);
b) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} – \frac{1}{{{{\sin }^2}x}}} \right)dx} \).
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Lời giải của Vua Trắc Nghiệm
a) Ta có \(\int\limits_0^{\frac{\pi }{2}} {\left( {3\cos x + 2\sin x} \right)dx} \) = \(\int\limits_0^{\frac{\pi }{2}} {3\cos xdx} + \int\limits_0^{\frac{\pi }{2}} {2\sin xdx} \)
= \(\left. {3\sin x} \right|_0^{^{\frac{\pi }{2}}} – \left. {2\cos x} \right|_0^{^{\frac{\pi }{2}}}\)
= \(3\sin \frac{\pi }{2} – 3\sin 0 – 2\cos \frac{\pi }{2} + 2\cos 0\)
= 5.
b) Ta có \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} – \frac{1}{{{{\sin }^2}x}}} \right)dx} \) = \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{1}{{{{\cos }^2}x}}dx} – \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{1}{{{{\sin }^2}x}}dx} \)
= \(\left. {\tan x} \right|_{_{\frac{\pi }{6}}}^{^{\frac{\pi }{4}}} – \left. {\cot x} \right|_{_{\frac{\pi }{6}}}^{^{\frac{\pi }{4}}}\)
= \(\tan \frac{\pi }{4} – \tan \frac{\pi }{6} – \cot \frac{\pi }{4} + \cot \frac{\pi }{6}\)
= 2 − \(\frac{{4\sqrt 3 }}{3}\).