Question Number 94574 by student work last updated on 19/May/20 | ||
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$$\int\frac{\sqrt{\mathrm{cosx}}}{\sqrt{\mathrm{sinx}\:}\:+\sqrt{\mathrm{cosx}}}\mathrm{dx}=? \\ $$ | ||
Commented by student work last updated on 19/May/20 | ||
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$$\mathrm{please}\:\mathrm{solve}\:\mathrm{who}\:\mathrm{can}? \\ $$ | ||
Answered by MJS last updated on 19/May/20 | ||
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$$\int\frac{\sqrt{\mathrm{cos}\:{x}}}{\sqrt{\mathrm{sin}\:{x}}+\sqrt{\mathrm{cos}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{dt}= \\ $$$$=\int\frac{{t}^{\mathrm{3}} −{t}^{\mathrm{2}} +{t}+\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}+\int\frac{{dt}}{{t}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}−\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\ $$ | ||
Commented by student work last updated on 20/May/20 | ||
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$$\:\:\int\frac{\mathrm{2t}}{\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt}=?\:\:\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{solve}? \\ $$ | ||
Commented by MJS last updated on 20/May/20 | ||
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$$\mathrm{just}\:\mathrm{decompose} \\ $$ | ||