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Question Number 163114 by abdullahhhhh last updated on 03/Jan/22

$$\int \left(\frac{\mathbf{cos}5\mathbf{x}+\mathbf{cos}4\mathbf{x}}{1+2\mathbf{cos}3\mathbf{x}}\right)\mathbf{dx}$$$$$$

Answered by blackmamba last updated on 04/Jan/22

$$\int \frac{\frac{e^{5x}+e^{-5x}}{2}+\frac{e^{4x}+e^{-4x}}{2}}{1+2\left(\frac{e^{3x}+e^{-3x}}{2}\right)}dx$$$$=\int \frac{e^{5x}+e^{-5x}+e^{4x}+e^{-4x}}{2+2e^{3x}+2e^{-3x}}dx$$$$let{e}^x=u\Rightarrow dx=\frac{du}{u}$$$$=\int \frac{u^5+u^{-5}+u^4+u^{-4}}{2+2u^3+2u^{-3}}\frac{du}{u}$$$$=\int \frac{u^4+u^{-6}+u^3+u^{-5}}{2+2u^3+2u^{-3}}du$$$$=\int \frac{u^{10}+1+u^9+u}{2u^6+2u^9+2u^3}du$$

Commented by mkam last updated on 04/Jan/22

$$false$$

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