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Category: Integration

Question-120059

Question Number 120059 by A8;15: last updated on 29/Oct/20 Answered by Lordose last updated on 29/Oct/20 Firstly,computetheindefiniteintegralI=xsec(2x)dx=14asec(a)da{a=2x}I=14(alntan(a2+π4)lntan(a2+π4)da){IBP}$$\mathrm{set}\:\frac{\mathrm{a}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\:=\:\mathrm{y}\:\Rightarrow\:\mathrm{da}=\mathrm{2dy}…

Question-119989

Question Number 119989 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xdx}}{\mathrm{2e}^{\mathrm{x}} −\mathrm{1}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xe}^{−\mathrm{x}} \mathrm{dx}}{\mathrm{2}−\mathrm{e}^{−\mathrm{x}}…

dx-x-2-25-x-2-

Question Number 119970 by bramlexs22 last updated on 28/Oct/20 dxx225x2? Answered by bemath last updated on 28/Oct/20 $$\:\int\:\frac{{dx}}{{x}^{\mathrm{3}} \sqrt{\frac{\mathrm{25}}{{x}^{\mathrm{2}} }−\mathrm{1}}}\:{dx}\:=\:\int\:\frac{{x}^{−\mathrm{3}} }{\:\sqrt{\mathrm{25}{x}^{−\mathrm{2}}…

Question-119960

Question Number 119960 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\left(\mathrm{1}+\mathrm{x}\right)\sqrt{\mathrm{x}}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}}…