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Question Number 65052 by AnjanDey last updated on 24/Jul/19

$$A.\mathrm{Evaluate}:$$$$\left(\mathrm{i}\right)\int \frac{\sin x+\cos x}{9+16\sin 2x}dx$$$$\left(\mathrm{ii}\right)\int \frac{1+x^2}{(1-x^2)\sqrt{1+x^2+x^4}}dx$$$$\left(\mathrm{iii}\right)\int \frac{x-1}{(x+1)\sqrt{x^3+x+x^2}}dx$$

Answered by MJS last updated on 24/Jul/19

$$\left(\mathrm{i}\right)$$$$\int \frac{\sin x+\cos x}{9+16\sin 2x}dx=$$$$[t=2\arctan x\to dx=\frac{2dt}{t^2+1}]$$$$=-2\int \frac{t^2-2t-1}{(t^2-8t+9)(9t^2+8t+1)}dt=$$$$=-18\int \frac{t^2-2t-1}{(t-4-\sqrt{7})(t-4+\sqrt{7})(9t+4-\sqrt{7})(9t+4+\sqrt{7})}dt=$$$$=-\frac{1}{40}\int \frac{dt}{t-4-\sqrt{7}}-\frac{1}{40}\int \frac{dt}{t-4+\sqrt{7}}+\frac{9}{40}\int \frac{dt}{9t+4-\sqrt{7}}+\frac{9}{40}\int \frac{dt}{9t+4+\sqrt{7}}=$$$$=-\frac{1}{40}\ln (t-4-\sqrt{7})-\frac{1}{40}\ln (t-4+\sqrt{7})+\frac{1}{40}\ln (9t+4-\sqrt{7})+\frac{1}{40}\ln (9t+4+\sqrt{7})=$$$$=\frac{1}{40}\ln \frac{81t^2+72t+9}{t^2-8t+9}=\frac{1}{40}\ln 9\frac{5-4\sqrt{2}\cos (x+\frac{\pi }{4})}{5+4\sqrt{2}\cos (x+\frac{\pi }{4})}=$$$$=\frac{1}{40}\ln \mid \frac{5-4\sqrt{2}\cos (x+\frac{\pi }{4})}{5+4\sqrt{2}\cos (x+\frac{\pi }{4})}\mid +C$$

Answered by MJS last updated on 24/Jul/19

$$\left(\mathrm{ii}\right)$$$$\int \frac{1+x^2}{(1-x^2)\sqrt{1+x^2+x^4}}dx=$$$$[t=2\arctan x\to dx=\frac{1+x^2}{2}dt]$$$$=\sqrt{2}\int \frac{dt}{\cos t\sqrt{7+\cos 2t}}=$$$$[u=\tan t\to dt={\cos }^{2}tdu]$$$$=\int \frac{du}{\sqrt{3u^2+4}}=\frac{\sqrt{3}}{3}\ln (\sqrt{3}u+\sqrt{3u^2+4})=$$$$=\frac{\sqrt{3}}{3}\ln \frac{2\sqrt{3}\sin t+\sqrt{14+2\cos 2t}}{2\cos t}=$$$$=\frac{\sqrt{3}}{3}\ln \left(2\frac{\sqrt{3}x+\sqrt{x^4+x^2+1}}{1-x^2}\right)=$$$$=\frac{\sqrt{3}}{3}\ln \mid \frac{\sqrt{3}x+\sqrt{x^4+x^2+1}}{1-x^2}\mid +C$$$$$$

Answered by Tanmay chaudhury last updated on 25/Jul/19

$$1)\int \frac{sinx+cosx}{9+16(1-1+sin2x)}$$$$\int \frac{d(sinx-cosx)}{9+16-16(1-sin2x)}$$$$\int \frac{d(sinx-cosx)}{25-16{(sinx-cosx)}^2}$$$$\frac{1}{16}\int \frac{d(sinx-cosx)}{{\left(\frac{5}{4}\right)}^2-{(sinx-cosx)}^2}$$$$\frac{1}{16}\times \frac{1}{2\times \frac{5}{4}}\times ln\left(\frac{\frac{5}{4}+(sinx-cosx)}{\frac{5}{4}-(sinx-cosx)}\right)+c$$$$$$$$$$$$$$$$$$$$$$

Answered by Tanmay chaudhury last updated on 25/Jul/19

$$ii)\int \frac{\frac{1}{x^2}+1}{(\frac{1}{x}-x)\sqrt{\frac{1}{x^2}+1+x^2}}dx$$$$\int \frac{d(x-\frac{1}{x})}{-(x-\frac{1}{x})\sqrt{{(x-\frac{1}{x})}^2+3}}$$$$x-\frac{1}{x}=\frac{1}{k}$$$$d(x-\frac{1}{x})=-\frac{1}{k^2}dk$$$$\int \frac{-dk}{k^2\times \frac{-1}{k}\times \sqrt{\frac{1}{k^2}+3}}$$$$\int \frac{dk}{\sqrt{1+3k^2}}$$$$\frac{1}{\sqrt{3}}\int \frac{dk}{\sqrt{k^2+\frac{1}{3}}}$$$$\frac{1}{\sqrt{3}}\times ln(k+\sqrt{k^2+\frac{1}{3}})+c$$$$\frac{1}{\sqrt{3}}ln[(x-\frac{1}{x})+\sqrt{{(x-\frac{1}{x})}^2+\frac{1}{3}}]+c$$$$$$$$$$$$$$

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