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Question Number 116391 by bobhans last updated on 03/Oct/20

$$\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{\mathrm{x}}}?$$

Commented by TANMAY PANACEA last updated on 03/Oct/20

$$howtopostquestion...$$$$hereiamsharingintregation...Tanmay$$$$\int \frac{sin\theta }{cos3\theta }+\frac{sin3\theta }{cos9\theta }+\frac{sin9\theta }{cos27\theta }d\theta$$

Commented by Dwaipayan Shikari last updated on 03/Oct/20

$$\mathrm{Tap}+\mathrm{to}\mathrm{post}\mathrm{question}$$

Commented by TANMAY PANACEA last updated on 03/Oct/20

$$okthankyousir$$

Commented by Dwaipayan Shikari last updated on 03/Oct/20

$$\int \frac{\sin \theta }{{4\cos }^3\theta -3\cos \theta }+\int \frac{\sin 3\theta }{{4\cos }^{3}3\theta -3\cos 3\theta }+\int \frac{\sin 9\theta }{{4\cos }^{3}9\theta -3\cos 9\theta }$$$$=\int \frac{-\mathrm{dt}}{{4\mathrm{t}}^3-3\mathrm{t}}-\frac{1}{3}\int \frac{\mathrm{du}}{{4\mathrm{u}}^3-3\mathrm{u}}-\frac{1}{9}\int \frac{\mathrm{dp}}{{4\mathrm{p}}^3-3\mathrm{p}}$$$$\Rightarrow \int \frac{-\mathrm{dt}}{\mathrm{t}({4\mathrm{t}}^2-3)}=\frac{1}{3}\int \frac{1}{\mathrm{t}}-\frac{4\mathrm{t}}{{4\mathrm{t}}^2-3}=\frac{1}{3}\log \left(\mathrm{t}\right)-\frac{1}{6}\log ({4\mathrm{t}}^2-3)$$$$\mathrm{And}\mathrm{similarly}$$$$-\frac{1}{3}\int \frac{\mathrm{du}}{{4\mathrm{u}}^3-3\mathrm{u}}=\frac{1}{9}\log \left(\mathrm{u}\right)-\frac{1}{18}\log ({4\mathrm{u}}^2-3)$$$$-\frac{1}{3}\int \frac{\mathrm{dp}}{{4\mathrm{p}}^3-3\mathrm{p}}=\frac{1}{27}\log \left(\mathrm{p}\right)-\frac{1}{54}\log ({4\mathrm{p}}^2-3)$$$$\mathrm{Integral}\mathrm{is}$$$$\frac{1}{3}\left(\log (\cos \theta .{\cos }^{\frac{1}{3}}\theta .{\cos }^{\frac{1}{9}}9\theta )\right)-\frac{1}{6}(\log \left(({4\cos }^2\theta -3){({4\cos }^{2}3\theta -3)}^{\frac{1}{3}}{({4\cos }^{2}9\theta -3)}^{\frac{1}{9}}\right)+\mathrm{C}$$

Commented by Dwaipayan Shikari last updated on 03/Oct/20

$$\mathrm{Kindly}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{tell}\mathrm{me}\mathrm{sir}.\mathrm{I}\mathrm{am}\mathrm{a}\mathrm{student}.\mathrm{Thanking}\mathrm{you}$$

Commented by mnjuly1970 last updated on 03/Oct/20

$$goodforyounicesolution$$$$peacebeuponyou...$$

Commented by TANMAY PANACEA last updated on 03/Oct/20

$$tan3\theta -tan\theta$$$$=\frac{sin3\theta cos\theta -sin\theta cos3\theta }{cos3\theta cos\theta }=\frac{2sin\theta cos\theta }{cos3\theta cos\theta }$$$$\frac{sin\theta }{cos3\theta }=\frac{1}{2}[tan3\theta -tan\theta ]$$$$\frac{sin3\theta }{cos9\theta }=\frac{1}{2}[tan9\theta -tan3\theta ]$$$$\frac{sin9\theta }{cos27\theta }=\frac{1}{2}[tan27\theta -tan9\theta ]$$$$addthem=\frac{1}{2}[tan27\theta -tan\theta ]$$$$soI=\int \frac{1}{2}(tan27\theta -tan\theta )d\theta$$$$=\frac{1}{2}[\frac{lnsec27\theta }{27}-lnsec\theta ]+c$$$$$$

Answered by bemath last updated on 03/Oct/20

$$\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{\mathrm{x}}}?$$$$\mathrm{Letting}\mathrm{x}={\mathrm{h}}^2\to \mathrm{dx}=2\mathrm{h}\mathrm{dh}$$$$\int \frac{2\mathrm{h}\mathrm{dh}}{({\mathrm{h}}^2+1).\mathrm{h}}=\int \frac{2\mathrm{dh}}{{\mathrm{h}}^2+1}=2{\tan }^{-1}\left(\mathrm{h}\right)+\mathrm{c}$$$$={2\tan }^{-1}\left(\sqrt{\mathrm{x}}\right)+\mathrm{c}$$

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