हिंदी

∫cos(logx) dx=F(x)+C where C is arbitrary constant. Here F(x) = -

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प्रश्न

`int cos(log x)  dx = F(x) + C` where C is arbitrary constant. Here F(x) =

विकल्प

  • `x[cos(log x) - sin (log (x))]`

  • `x/2 [cos(log x) + sin (log (x))]`

  • `x[cos(log x) + sin (log (x))]`

  • `x/2 [cos(log x) - sin (log (x))]`

MCQ

उत्तर

`x/2 [cos(log x) + sin (log (x))]`

Explanation:

`I - int cos(log x). 1  dx`

= `cos (log x) int 1  dx - int ((d cos(log  x))/(dx) int 1 dx) dx`

= `x cos (log + x) + int sin(log x). 1/x. x  dx`

= `x cos (log  x) + sin  (log  x) int 1  dx - int ((d sin(log  x))/(dx) int 1  dx) dx`

= `x cos (log x) + x sin (log x) - int cos(log x) dx`

`I = x [cos (log x) + sin (log x)] - I + C_1`

`2I = x[cos,(log x) + sin (log  x)] + C_1`

`I = x/2 [cos (log x) + sin (log x)] + C`

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