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

Question

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

`int_(sin theta)^(cos theta) f(x tan theta)dx` (where `theta!=(npi)/2,n epsilonI`) is equal to
A. `-cos theta int_(1)^(tan theta) f(x sin theta )dx`
B. `-tan theta int_(cos theta)^(sin theta) f(x)dx`
C. `sin theta int_(1)^(tan theta)f (x cos theta)dx`
D. `1/(tan theta) int_(sin theta)^(sin theta tan theta) f(x)dx`

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
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`

7 views Answered 2023-02-05 15:29