1.

`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`

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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