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Category: Integration

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Question Number 118338 by benjo_mathlover last updated on 17/Oct/20 solvedx35sinx Answered by Dwaipayan Shikari last updated on 17/Oct/20 dx35sinx$$=\mathrm{2}\int\frac{{dt}}{\mathrm{3}−\frac{\mathrm{10}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}}…

Question-183865

Question Number 183865 by Michaelfaraday last updated on 31/Dec/22 Answered by MJS_new last updated on 31/Dec/22 =π012sinx2dx=$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{3}} {\int}}\left(\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\:−\mathrm{1}\right){dx}+\underset{\pi/\mathrm{3}} {\overset{\pi} {\int}}\left(\mathrm{1}−\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\right){dx}=…

Question-183844

Question Number 183844 by Michaelfaraday last updated on 30/Dec/22 Answered by MJS_new last updated on 31/Dec/22 dx2x1+x3==2x1x3(2x1+x3)(2x1x3)dx==2x1x+2dxx3x+2dx=withu=2x1andv=x3itseasytoget$$=\mathrm{2}\sqrt{\mathrm{2}{x}−\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{5}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\:\sqrt{\mathrm{5}}}\:−\mathrm{2}\sqrt{{x}−\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{5}}\mathrm{arctan}\:\frac{\sqrt{{x}−\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\:+{C}…

Question-183845

Question Number 183845 by Michaelfaraday last updated on 30/Dec/22 Answered by MJS_new last updated on 31/Dec/22 dx(x+1)x2+4x+2=[t=x+2+x2+4x+22dx=x2+4x+2tdt]$$=\mathrm{2}\int\frac{{dt}}{\:\sqrt{\mathrm{2}}{t}^{\mathrm{2}} −\mathrm{2}{t}+\sqrt{\mathrm{2}}}=\mathrm{2arctan}\:\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)\:=…