InterviewSolution
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InterviewSolution
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Evaluate : (i) `int(sinx)/(sqrt(4cos^(2)x-1))dx` (ii) `int(sec^(2)x)/(sqrt(tan^(2)x-4))dx` |
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Answer» (i) Putting `cosx=t and - sin x dx = dt` , we get `int(sinx)/(sqrt(4 cos^(2)x-1))dx=int(-dt)/(sqrt(4t^(2)-1))=-(1)/(2)int(dt)/(sqrt(t^(2)-(1//2))^(2))` `=-(1)/(2)*log|t+sqrt(t^(2)-(1)/(4))|+C` `=-(1)/(2)*log|2t+sqrt(4t^(2)-1)|+C` `=-(1)/(2)log|2cosx+sqrt(4cos^(2)x-1)|+C`. Putting `tanx=t and sec^(2)x dx = dt` , we get `int(sec^(2)x)/(sqrt(tan^(2)x-4))dx=int(dt)/(sqrt(t^(2)-4))=log|t+sqrt(t^(2)-4)|+C` `=log|tanx+sqrt(tan^(2)x-4)|+C`. |
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