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Question Number 186527 by Mingma last updated on 05/Feb/23

Answered by ARUNG_Brandon_MBU last updated on 05/Feb/23

$$I=\int \frac{\cos x}{{\sin }^{2}x+\sin x+1}dx,s=\sin x$$$$=\int \frac{ds}{s^2+s+1}=\int \frac{ds}{{(s+\frac{1}{2})}^2+\frac{3}{4}}$$$$=\frac{2}{\sqrt{3}}\arctan \left(\frac{2s+1}{\sqrt{3}}\right)+C$$$$=\frac{2\sqrt{3}}{3}\arctan \left(\frac{2\sin x+1}{\sqrt{3}}\right)+C$$

Commented by aba last updated on 05/Feb/23

$$\int \frac{1}{{\mathrm{u}}^2+{\mathrm{a}}^2}\mathrm{du}=\frac{1}{\mathrm{a}}\arctan \left(\frac{\mathrm{u}}{\mathrm{a}}\right)+\mathrm{C}$$

Commented by ARUNG_Brandon_MBU last updated on 05/Feb/23

Thanks.

Answered by aba last updated on 05/Feb/23

$$\mathrm{let}\mathrm{t}=\sin \left(\mathrm{x}\right)\Rightarrow \mathrm{dt}=\cos \left(\mathrm{x}\right)\mathrm{dx}$$$$\int \frac{\cos \left(\mathrm{x}\right)\mathrm{dx}}{{\sin }^2\left(\mathrm{x}\right)+\sin \left(\mathrm{x}\right)+1}=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+\mathrm{t}+1}=\int \frac{\mathrm{dt}}{{(\mathrm{t}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{2}{\sqrt{3}}\arctan \left(\frac{\mathrm{t}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\mathrm{C}=\frac{2}{\sqrt{3}}\arctan \left(\frac{2\sin \left(\mathrm{x}\right)+1}{\sqrt{3}}\right)+\mathrm{C}$$$$$$

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