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Category: Integration

cos-9x-cos-4x-cos-2x-dx-

Question Number 187251 by horsebrand11 last updated on 15/Feb/23 $$\:\:\int\:\frac{\mathrm{cos}\:\mathrm{9}{x}}{\mathrm{cos}\:\mathrm{4}{x}.\:\mathrm{cos}\:\mathrm{2}{x}}\:{dx}=? \\ $$ Commented by MJS_new last updated on 15/Feb/23 $$=\mathrm{4}\int\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \:{x}\right)\mathrm{cos}\:{x}\:{dx}− \\ $$$$\:\:\:\:−\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}}{dx}− \\…

Question-121712

Question Number 121712 by rs4089 last updated on 11/Nov/20 Answered by Ar Brandon last updated on 11/Nov/20 $$\int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}}} \int_{\mathrm{y}} ^{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…

Question-121704

Question Number 121704 by rs4089 last updated on 11/Nov/20 Answered by bemath last updated on 11/Nov/20 $$\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mid\:\left(\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)\mid_{\mathrm{1}} ^{\mathrm{2}} \:{dy}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\left(\mathrm{2}{y}^{\mathrm{2}}…

x-4-5x-3-6x-2-18-x-3-3x-2-dx-

Question Number 121696 by bemath last updated on 11/Nov/20 $$\:\:\int\:\frac{{x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{18}}{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} }\:{dx}\:? \\ $$ Answered by liberty last updated on 11/Nov/20 $$\mathrm{J}=\int\:\left[\left(\mathrm{x}−\mathrm{2}\right)−\frac{\mathrm{18}}{\mathrm{x}^{\mathrm{2}}…

advanced-calculus-prove-that-0-sin-4-x-ln-x-x-2-dx-pi-4-1-euler-mascheroni-constant-m-n-july-1970-

Question Number 121674 by mnjuly1970 last updated on 10/Nov/20 $$\:\:\:\:\:\:\:\:\:…\:\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:{prove}\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{4}} \left({x}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−\gamma\right) \\ $$$$\gamma:{euler}−{mascheroni}\:{constant} \\ $$$$\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970} \\…

Question-187164

Question Number 187164 by mathlove last updated on 14/Feb/23 Answered by qaz last updated on 14/Feb/23 $$\underset{{p}\rightarrow{ln}\frac{\mathrm{1}}{\mathrm{2}}} {{lim}}\frac{{ln}\mathrm{2}+{ln}\mathrm{2}\:{cos}\:{p}}{\left({cos}\:{ln}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=\frac{\left(\mathrm{1}+{cos}\:{ln}\mathrm{2}\right){ln}\mathrm{2}}{{cos}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}}=\frac{\left[\mathrm{2cos}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}\right]{ln}\mathrm{2}}{{cos}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}}=\mathrm{2}{ln}\mathrm{2} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty}…