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`int cos (log_e x)dx` is equal toA. `(1)/(2)xx[cos(log_ex)+sin(log_ex)]`B. `(x)[cos(log_e x)+sin(log_ex)]`C. `(1)/(2)x[cos(log_e x)-sin(log_e x)]`D. `x[cos (log_ex)-sin(log_ex)]` |
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Answer» Correct Answer - A Let `l=intcos(log_ex)dx` `=cos(log_eX)x-int((-sinlog_ex)/x)xdx` `= x cos (log_ex)+int sin(log_ex).1dx` `=xcos(log_ex)+xsin(log_ex)-int(cos(log_ex))/(x)x dx` `=xcos(log_ex)+ xsin(log_ex)-l` `rArr 2l=x[cosIog_ex)+sin(log_ex)]` `rArr l=(x)/(2)[cos(log_ex)+sin(log_ex)]` |
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