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Question

$\int\limits_{0}^{\frac{\pi }{4}}\frac{cos(x-\frac{\pi }{4})}{4-3cosx}dx$

binhnguyenhoangvu · Answer 1 · 5/14/2013 12:11:32 AM

Mình tính nguyên hàm, bạn tự thế cận nhé!

Tính

$I= \int {\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}}} dx$

Ta có:

$\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}} = \frac{{\sqrt 2 }}{2}.\frac{{\sin x + \cos x}}{{4 - 3\cos x}}$

$= \frac{{\sqrt 2 }}{2}.\frac{{\sin x}}{{4 - 3\cos x}} + \frac{{\sqrt 2 }}{2}\left( { - \frac{1}{3} + \frac{4}{3}.\frac{1}{{4 - 3\cos x}}} \right)$

$\Rightarrow I = \frac{{\sqrt 2 }}{2}\int {\frac{{\sin x}}{{4 - 3\cos x}}} dx - \frac{{\sqrt 2 }}{6}\int {dx} + \frac{{2\sqrt 2 }}{3}\int {\frac{1}{{4 - 3\cos x}}} dx$

* Tính

$A = \int {\frac{{\sin x}}{{4 - 3\cos x}}} dx$ : đặt

$t = 4 - 3\cos x$

* Tính

$B = \int {\frac{1}{{4 - 3\cos x}}} dx$

$4 - 3\cos x = 4({\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2}) - 3({\cos ^2}\frac{x}{2} - {\sin ^2}\frac{x}{2}) = 7{\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2} = {\cos ^2}\frac{x}{2}(7{\tan ^2}\frac{x}{2} + 1)$