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Category: Integration

Question-159823

Question Number 159823 by tounghoungko last updated on 21/Nov/21 Answered by Ar Brandon last updated on 21/Nov/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}}\mathrm{sin}^{−\mathrm{1}} {xdx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mathrm{sin}^{−\mathrm{1}} {xdx}…

Question-94269

Question Number 94269 by LPM last updated on 17/May/20 Commented by prakash jain last updated on 17/May/20 $$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}}…

By-using-the-double-integral-find-the-area-of-the-regiin-bounded-by-the-curve-y-x-2-2x-y-x-2-2x-and-x-axis-sketch-the-region-of-integration-pleas-sir-help-me-

Question Number 94243 by mhmd last updated on 17/May/20 $${By}\:{using}\:{the}\:{double}\:{integral}\:{find}\:{the}\:{area}\:{of}\:{the}\:{regiin}\: \\ $$$${bounded}\:{by}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} +\mathrm{2}{x}\:,\:{y}={x}^{\mathrm{2}} −\mathrm{2}{x}\:{and}\:{x}−{axis}\: \\ $$$$\left({sketch}\:{the}\:{region}\:{of}\:{integration}\right) \\ $$$${pleas}\:{sir}\:{help}\:{me}\: \\ $$ Commented by Ar Brandon last…

solve-the-integrayion-sin2x-sin5xsin3x-

Question Number 28705 by students last updated on 29/Jan/18 $${solve}\:{the}\:{integrayion}\:\frac{{sin}\mathrm{2}{x}}{{sin}\mathrm{5}{xsin}\mathrm{3}{x}} \\ $$ Answered by mrW2 last updated on 30/Jan/18 $$\mathrm{sin}\:\mathrm{2}{x}=\mathrm{sin}\:\left(\mathrm{5}{x}−\mathrm{3}{x}\right)=\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{5}{x}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\int\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{sin}\:\mathrm{3}{x}}{dx}=\int\mathrm{cot}\:\mathrm{3}{x}\:{dx}−\int\mathrm{cot}\:\mathrm{5}{x}\:{dx} \\ $$$$=\frac{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{3}{x}\right)}{\mathrm{3}}−\frac{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{5}{x}\right)}{\mathrm{5}}+{C} \\…