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`inttanxdx=?`A. `log|cosx|+C`B. `-log|cosx|+C`C. `log|sinx|+C`D. `-log|sinx|+C` |
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Answer» Correct Answer - B `I=int(sinx)/(cosx)dx=int(-dt)/(t)`, where cos x=t. `=-log|t|+C=-log|cosx|+C`. |
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