InterviewSolution
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InterviewSolution
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Evaluate : (i) `int(sinx)/((1-4cos^(2)x))dx` (ii) `int("cosec"^(2)x)/((1-cot^(2)x))dx` |
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Answer» (i) Putting `cosx=t" and "-sinx dx=dt`, we get `int(sinx)/((1-4cos^(2)x))dx=-int(dt)/((1-4t^(2)))` `=-(1)/(4)*int(dt)/(((1)/(4)-t^(2)))=(1)/(4)*int(dt)/({((1)/(2))^(2)-t^(2)})` `=-(1)/(4)xx(1)/((2xx(1)/(2)))log|((1)/(2)+t)/((1)/(2)-t)|+C` `=-(1)/(4)log|(1+2t)/(1-2t)|+C=-(1)/(4)log|(1+2cosx)/(1-cosx)|+C`. (ii) Putting `cotx=t" and -cosec"^(2)xdx=dt` we get `int("cosec"^(2)x)/((1-cot^(2)x))dx=-int(dt)/((1-t^(2)))=-(1)/((2xx1))log|(1+t)/(1-t)|+C` `=-(1)/(2)log|(1+cotx)/(1-cotx)|+C`. |
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