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Category: Integration

After-looking-at-a-previous-question-I-was-wondering-if-the-following-is-correct-I-n-0-n-1-x-dx-n-R-I-n-0-n-e-ipix-dx-1-I-n-i-pi-e-ipix-0-n-I-n-i-pi-e-ipix-0-n-

Question Number 6441 by Temp last updated on 27/Jun/16 $$\mathrm{After}\:\mathrm{looking}\:\mathrm{at}\:\mathrm{a}\:\mathrm{previous}\:\mathrm{question} \\ $$$$\mathrm{I}\:\mathrm{was}\:\mathrm{wondering}\:\mathrm{if}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{is}\:\mathrm{correct}: \\ $$$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\:{n}} \left(−\mathrm{1}\right)^{{x}} {dx},\:\:{n}\in\mathbb{R} \\ $$$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\:{n}} {e}^{{i}\pi{x}} {dx}\:\:\:\left(\mathrm{1}\right) \\…

I-dx-sinx-sin2x-

Question Number 6444 by sanusihammed last updated on 27/Jun/16 $${I}\:=\:\int\frac{{dx}}{{sinx}\:+\:{sin}\mathrm{2}{x}} \\ $$ Commented by Temp last updated on 27/Jun/16 $$\mathrm{Try}\:\mathrm{www}.\mathrm{WolframAlpha}.\mathrm{com} \\ $$$$\mathrm{search}: \\ $$$$“{int}\:\mathrm{1}/\left({sin}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)\right)\:{dx}'' \\…

H-0-4-tanx-dx-

Question Number 6439 by sanusihammed last updated on 27/Jun/16 $${H}\:=\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \sqrt{{tanx}}\:{dx}\: \\ $$ Commented by Temp last updated on 27/Jun/16 $$\int\sqrt{\mathrm{tan}{x}}{dx}\:\mathrm{is}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{solve}. \\ $$ Commented…

sin-1-x-x-3-dx-

Question Number 137508 by EDWIN88 last updated on 03/Apr/21 $$\int\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{3}} }\:{dx}\:=? \\ $$ Answered by liberty last updated on 03/Apr/21 $${L}=\int\:\frac{\mathrm{1}}{{x}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:{dx} \\ $$$${let}\:{t}\:=\:\frac{\mathrm{1}}{{x}}\:\Rightarrow−{dt}\:=\:\frac{{dx}}{{x}^{\mathrm{2}} }…

advanced-calculus-prove-that-0-1-ln-x-ln-1-x-1-x-dx-13-8-3-pi-2-4-ln-2-

Question Number 137501 by mnjuly1970 last updated on 03/Apr/21 $$\:\:\:\:\:\:\:….{advanced}\:….\:{calculus}…. \\ $$$$\:\:{prove}\:{that}:: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right).{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx}=\frac{\mathrm{13}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\left(\mathrm{2}\right)…. \\ $$ Answered by Ñï= last updated on…