Question

Tính các tích phân sau:

a) \(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{6}}\dfrac{1}{\sin^2x}dx;\)                       b) \(\int\limits^{\dfrac{\pi}{4}}_0\left(1+\tan x\right)\cos xdx\).

Answer 1

a) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{1}{{{{\sin }^2}x}}dx}  = \left. {\left( { - \cot x} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{4}} = \left( { - \cot \frac{\pi }{4}} \right) - \left( { - \cot \frac{\pi }{6}} \right) =  - 1 + \sqrt 3 \)

b) \(\int\limits_0^{\frac{\pi }{4}} {\left( {1 + \tan x} \right)\cos xdx}  = \int\limits_0^{\frac{\pi }{4}} {\left( {\cos x + \sin x} \right)dx}  = \left. {\left( {\sin x - \cos x} \right)} \right|_0^{\frac{\pi }{4}}\)

\( = \left( {\sin \frac{\pi }{4} - \cos \frac{\pi }{4}} \right) - \left( {\sin 0 - \cos 0} \right) = 0 - \left( { - 1} \right) = 1\)