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Category: Integration

Question-219384

Question Number 219384 by mnjuly1970 last updated on 23/Apr/25 Answered by SdC355 last updated on 24/Apr/25 $$\int\:\:\:\frac{\mathrm{d}{x}}{\:\sqrt{{x}}}\:\mathrm{cos}^{\mathrm{3}} \left({x}\right)\mathrm{sin}\left({x}\right)={I} \\ $$$$\int\:\:\:\frac{\mathrm{d}{x}}{\:{x}}\:\sqrt{{x}}\mathrm{cos}^{\mathrm{3}} \left({x}\right)\mathrm{sin}\left({x}\right)=\int\:\:\:\frac{\mathrm{d}{x}}{{x}}\:\sqrt{{x}}\mathrm{cos}^{\mathrm{2}} \left({x}\right)\mathrm{cos}\left({x}\right)\mathrm{sin}\left({x}\right) \\ $$$$\int\:\:\frac{\mathrm{d}{x}}{\:\mathrm{2}{x}}\:\sqrt{{x}}\mathrm{cos}^{\mathrm{2}} \left({x}\right)\mathrm{sin}\left(\mathrm{2}{x}\right)=\int\:\:\mathrm{d}{x}\:\frac{{e}^{−{xt}}…

Prove-I-0-x-1-pi-0-pi-e-x-cox-d-x-2-I-0-x-xI-0-x-x-2-I-0-x-0-

Question Number 219305 by Nicholas666 last updated on 22/Apr/25 $$ \\ $$$$\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:{I}_{\mathrm{0}} \left({x}\right)\:=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\:{x}\:{cox}\left(\theta\right)} \:{d}\theta\:; \\ $$$$\:\:\:{x}^{\mathrm{2}} {I}_{\mathrm{0}} ^{''} \left({x}\right)\:+\:{xI}'_{\mathrm{0}} \left({x}\right)\:−\:{x}^{\mathrm{2}} {I}_{\mathrm{0}}…

x-1-x-2-1-x-3-dx-

Question Number 219193 by fantastic last updated on 20/Apr/25 $$\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}\:.\frac{\mathrm{1}}{{x}+\mathrm{3}}\:{dx}=? \\ $$ Answered by aleks041103 last updated on 20/Apr/25 $$\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}={u}=\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{1}−{u}}−\mathrm{2}=\frac{\mathrm{2}{u}−\mathrm{1}}{\mathrm{1}−{u}} \\ $$$$\Rightarrow{dx}=\frac{\mathrm{2}\left(\mathrm{1}−{u}\right)−\left(−\mathrm{1}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }{du}=…