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Category: Integration

asin-2-x-bcos-2-x-dx-

Question Number 20467 by tammi last updated on 27/Aug/17 (asin2x+bcos2x)dx Answered by ajfour last updated on 27/Aug/17 =a2(1cos2x)dx+b2(1+cos2x)dx$$=\frac{{a}}{\mathrm{2}}\left({x}−\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\right)+\frac{{b}}{\mathrm{2}}\left({x}+\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\right)+{C} \

cos-2xdx-sin-2-2x-8-

Question Number 20465 by tammi last updated on 27/Aug/17 cos2xdxsin22x+8 Answered by sma3l2996 last updated on 27/Aug/17 t=sin(2x)dt=2cos(2x)dx$$\int\frac{{cos}\mathrm{2}{xdx}}{{sin}^{\mathrm{2}} \mathrm{2}{x}+\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{8}\left(\left(\frac{{t}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)^{\mathrm{2}}…