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The integral `intcos(log_(e)x)dx` is equal to: (where C is a constant of integration)A. `(x)/(2)[cos(log_(e)x)+sin(log_(e)x)]+C`B. `x[cos(log_(e)x)+sin(log_(e)x)]+C`C. `x[cos(log_(e)x)-sin(log_(e)x)]+C`D. `(x)/(2)[cos(log_(e)x)-sin(log_(e)x)]+C` |
Answer» Correct Answer - A Let `I=int cos(log_(e)x)dx` `=x cos(log_(e)x)-intx(-sin(log_(e)x))(1)/(x)*dx" "["using integration by parts"]` `= x cos(log_(e)x)+int sin(log_(e)x)dx` `=x cos(log_(e)x)+ x sin(log_(e)x)-int x(cos(log_(e)x))(1)/(x)dx" "["again, using integration by parts"]` `rArr I = x cos(log_(e)x)+x sin(log_(e)x)-I` `rArr I=(x)/(2)[cos(log_(e)x)+sin(log_(e)x)]+C`. |
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