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Category: Integration

Question-200606

Question Number 200606 by Calculusboy last updated on 20/Nov/23 Answered by Frix last updated on 20/Nov/23 cosxsin2xcosx+sinxdx=t=xπ4=(14+costsint2sin2t2tant4)dt=$$=\frac{{t}}{\mathrm{4}}−\frac{\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{4}}−\frac{{t}+\mathrm{cos}\:{t}\:\mathrm{sin}\:{t}}{\mathrm{4}}+\frac{\mathrm{ln}\:\mathrm{cos}\:{t}}{\mathrm{4}}=…

Question-200602

Question Number 200602 by Calculusboy last updated on 20/Nov/23 Answered by Frix last updated on 21/Nov/23 Useu=11+costu=tant2;v=t+sintv=1+costt+sint1+costdt==(t+sint)tant2(1+cost)tant2dt=$$={t}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:+\mathrm{1}−\mathrm{cos}\:{t}\:−\int\mathrm{sin}\:{t}\:{dt}=…

Question-200586

Question Number 200586 by Calculusboy last updated on 20/Nov/23 Answered by Frix last updated on 21/Nov/23 $$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{2}+\mathrm{tan}^{\mathrm{2}} \:{x}}{dx}=\underset{\mathrm{0}} {\overset{\pi} {\int}}{xdx}−\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}…

Question-200444

Question Number 200444 by Calculusboy last updated on 18/Nov/23 Answered by Frix last updated on 19/Nov/23 eix2dx=t=eiπ4x$$=\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{2}}\int\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{4}}\int\frac{\mathrm{2e}^{{t}^{\mathrm{3}}…