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`int(1)/(sinx*cos^(2)x)dx=`A. `secx+log|secx+tanx|+C`B. `secx+log|secx+tanx|+C`C. `secx+log|secx-tanx|+C`D. `secx+log|"cosec"x-cotx|+C` |
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Answer» Correct Answer - D We have, `l=int(dx)/(sinx*cos^(2)x)` `rArr" "l=int(sin^(2)x+cos^(2)x)/(sinxcos^(2)x)dx` `rArrl=int(sin^(2)x)/(sinxcos^(2)x)dx+int(cos^(2)x)/(sinxcos^(2)x)dx` `rArrl=intsecxtanxdx+int"cosec"xdx` `rArrl=secx+log|"cosec"x-cotx|+C` |
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