int(sin theta)^(cos theta) f(x tan theta...

`int_(sin theta)^(cos theta) f(x tan theta)dx` (where `theta!=(npi)/2,n in I`) is equal to

`-cos theta int_(1)^(tan theta) f(x sin theta )dx`

`-tan theta int_(cos theta)^(sin theta) f(x)dx`

`sin theta int_(1)^(tan theta)f (x cos theta)dx`

`-1/(tan theta) int_(sin theta)^(sin theta tan theta) f(x)dx`

Text Solution

Verified by Experts

The correct Answer is:

Putting `x tan theta =z sin theta` or `dx=cos theta dz`. We get
`I=cos theta int_(tan theta)^(1) f(z sin theta)dz`
`=-cos theta int_(1)^(tan theta)f(x sin theta) dx`

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