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Question Number 83719 by Roland Mbunwe last updated on 05/Mar/20

$$Question.\stackrel{Showthat{\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{cosx}{3+{cos}^{2}x}dx=\frac{1}{4}ln3}{}$$

Commented by mathmax by abdo last updated on 05/Mar/20

$$I={\int }_0^{\frac{\pi }{2}}\frac{cosx}{3+{cos}^{2}x}dx\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{cosxdx}{4-{sin}^{2}x}{=}_{sinx=t}{\int }_0^1\frac{dt}{4-t^2}$$$$=\frac{1}{4}{\int }_0^1(\frac{1}{2-t}+\frac{1}{2+t})dt=\frac{1}{4}{[ln\mid \frac{2+t}{2-t}\mid ]}_0^1=\frac{1}{4}\left(ln\left(3\right)\right)=\frac{1}{4}ln\left(3\right)$$

Answered by jagoll last updated on 05/Mar/20

$$\int \frac{\cos \mathrm{x}\mathrm{dx}}{4-\sin {}^2\mathrm{x}}=\int \frac{\mathrm{d}\left(\sin \mathrm{x}\right)}{4-\sin {}^2\mathrm{x}}$$$$\mathrm{let}\mathrm{u}=\sin \mathrm{x}$$$$\int \frac{\mathrm{du}}{(2+\mathrm{u})(2-\mathrm{u})}=\int \frac{1}{4}\frac{\mathrm{du}}{(2-\mathrm{u})}+\int \frac{1}{4}\frac{\mathrm{du}}{(2+\mathrm{u})}$$$$=\frac{1}{4}\ln \mid 4-{\mathrm{u}}^2\mid +\mathrm{c}$$$$=\frac{1}{4}\ln \mid 4-\sin {}^2\mathrm{x}\mid +\mathrm{c}$$$$=\frac{1}{4}\ln \mid 3+\cos {}^2\mathrm{x}\mid {\mid }_0^{\frac{\pi }{2}}$$$$=\frac{1}{4}\ln \mid 3\mid -\frac{1}{4}\ln \mid 4\mid$$

Commented by john santu last updated on 05/Mar/20

$$\mathrm{typo}\mathrm{sir}.$$$$\int \frac{1}{4}\frac{\mathrm{du}}{2-\mathrm{u}}=-\frac{1}{4}\ln \mid 2-\mathrm{u}\mid$$$$\mathrm{so}=-\frac{1}{4}\ln \mid 2-\mathrm{u}\mid +\frac{1}{4}\ln \mid 2+\mathrm{u}\mid$$$$=\frac{1}{4}\ln \mid \frac{2+\sin \mathrm{x}}{2-\sin \mathrm{x}}\mid +\mathrm{c}$$

Answered by MJS last updated on 05/Mar/20

$$\int \frac{\cos x}{3+{\cos }^{2}x}dx=$$$$[t=\sin x\to \frac{dt}{\cos x}]$$$$=-\int \frac{dt}{t^2-4}=-\frac{1}{4}\ln \mid \frac{t-2}{t+2}\mid =$$$$=-\frac{1}{4}\ln \mid \frac{\sin x-2}{\sin x+2}\mid +C$$$$\Rightarrow \underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\cos x}{3+{\cos }^{2}x}dx=\frac{1}{4}\ln 3$$

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