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`int (sin^2x cos^2x)/(sin^5x+cos^3x sin^2x + sin^3x cos^2x + cos^5x)^2 dx`A. `(1)/(3(1+tan^(3)x))+C`B. `(-1)/(3(1+tan^(3)x))+C`C. `(1)/(1+cot^(3)x)+C`D. `(-1)/(1+cot^(3)x)+C` |
Answer» Correct Answer - B We have, `I = int(sin^(2)x*cos^(2)x)/((sin^(6)x + cos^(3)x*sin^(2)x+sin^(3)x*cos^(2)x+cos^(5)x)^(2))dx` ` = int(sin^(2)x cos^(2)x)/{{sin^(3)x(sin^(2) x + cos^(2)x)+cos^(3)x(sin^(2)x +cos^(2)x)}^(2))dx` `=int(sin^(2)x cos^(2) x)/((sin^(3)x + cos^(3)x)^(2))dx=int(sin^(2)x cos^(2)x)/(cos^(6)x(1+tan^(3)x)^(2))dx` `=int(tan^(2)x sec^(2)x)/((1+tan^(3)x)^(2))dx` Put `tan^(3) x = t rArr 3 tan^(2)x sec^(2) xdx = dt` `therefore" "I = (1)/(3)int(dt)/((1+t)^(2))` `rArr" "I = (-1)/(3(1+t))+C rArr I=(-1)/(3(1+tan^(3)x))+C` |
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