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




Question Number 20467 by tammi last updated on 27/Aug/17
∫(asin^2 x+bcos^2 x)dx
$$\int\left({a}\mathrm{sin}\:^{\mathrm{2}} {x}+{b}\mathrm{cos}\:^{\mathrm{2}} {x}\right){dx} \\ $$
Answered by ajfour last updated on 27/Aug/17
=(a/2)∫(1−cos 2x)dx+(b/2)∫(1+cos 2x)dx  =(a/2)(x−((sin 2x)/2))+(b/2)(x+((sin 2x)/2))+C  =(((a+b)/2))x−(((a−b)/2))((sin 2x)/2) +C .
$$=\frac{{a}}{\mathrm{2}}\int\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right){dx}+\frac{{b}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right){dx} \\ $$$$=\frac{{a}}{\mathrm{2}}\left({x}−\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\right)+\frac{{b}}{\mathrm{2}}\left({x}+\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\right)+{C} \\ $$$$=\left(\frac{{a}+{b}}{\mathrm{2}}\right){x}−\left(\frac{{a}−{b}}{\mathrm{2}}\right)\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\:+{C}\:. \\ $$

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