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Question Number 62732 by mathmax by abdo last updated on 24/Jun/19

$$calculate{\int }_0^{2\pi }\frac{cos\left(2x\right)}{2cosx-sin\left(x\right)}dx$$

Answered by MJS last updated on 24/Jun/19

$$\frac{\cos \left(2(x+\pi )\right)}{2\cos (x+\pi )-\sin (x+\pi )}=-\frac{\cos 2x}{2\cos x-\sin x}\Rightarrow$$$$\Rightarrow \underset{0}{\stackrel{\pi }{\int }}\frac{\cos 2x}{2\cos x-\sin x}dx=-\underset{\pi }{\stackrel{2\pi }{\int }}\frac{\cos 2x}{2\cos x-\sin x}dx\Rightarrow$$$$\Rightarrow \underset{0}{\stackrel{2\pi }{\int }}\frac{\cos 2x}{2\cos x-\sin x}dx=0$$

Commented by mathmax by abdo last updated on 24/Jun/19

$$thankyousir.$$

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