1.

`inte^(-x)(1-tanx)secx dx` is equal toA. `e^(x)cos x+C`B. `e^(x)sec x+C`C. `e^(x)sinx+C`D. `e^(x)tanx+C`

Answer» Correct Answer - B
Let`" "l=int e^(x) sec x (1+tanx)dx`
`rArr" "l=int e^(x)sec x dx+int e^(x)sec x tanx dx" …(i)"`
Now, `int e^(x)sec x dx`
On applying integration by parts, we get
`=secx inte^(x) dx-int[(d)/(dx)secx int e^(x)dx]dx`
`=e^(x)sec x-int sec x tanx e^(x)dx" ...(ii)"`
`l=e^(x) sec x - int e^(x)secx tanx dx+ int sec x tan x e^(x)dx+C`
`rArr" "l=e^(x)sec x +C`


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