InterviewSolution
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`(1)/(cos(x+a)cos(x+b))` |
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Answer» `int(1)/(cos(x+a)cos(x+b))dx=I" "` (माना ) `thereforeI=(1)/(sin(a-b))int(sin(a-b))/(cos(x+a)cos(x+b))dx` `rArr I=(1)/(sin(a-b))int(sin[(x+a)-(x+b)])/(cos(x+a)cos(x+b))dx sin(x+a)cos(x+b)` `=(1)/(sin(a-b))int(-cos(x+a)sin(x_+b))/(cos(x+a)cos(x+b))dx` `=(1)/(sin(a-b))int[(sin(x+a))/(cos(x+a))-(sin(x+b))/(cos(x+b))]dx` `=(1)/(sin(a-b))int[tan(x+a)-tan(x+b)]dx` `=(1)/(sin(a-b))[-log|cos(x+a)|+log|cos(x+b)|]+C` `=(1)/(sin(a-b))log|(cos(x+b))/(cos(x+a))|+C` |
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