The value of the definite integral `int_0^(pi/2)sqrt(tanx)dx` is `sqrt(2)pi` (b) `pi/(sqrt(2))` `2sqrt(2)pi` (d) `pi/(2sqrt(2))`

Question

The value of the definite integral `int_0^(pi/2)sqrt(tanx)dx` is `sqrt(2)pi` (b) `pi/(sqrt(2))` `2sqrt(2)pi` (d) `pi/(2sqrt(2))`

The value of the definite integral `int_0^(pi/2)sqrt(tanx)dx` is `sqrt(2)pi` (b) `pi/(sqrt(2))` `2sqrt(2)pi` (d) `pi/(2sqrt(2))`
A. `sqrt(2)pi`
B. `(pi)/(sqrt(2))`
C. `2sqrt(2)pi`
D. `(pi)/(2sqrt(2))`

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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

Correct Answer - B
`I=int_(0)^(pi//2) sqrt(tanx) dx`…………….1
or `I=int_(0)^(pi//2) sqrt(cotx)dx`…………….2
Adding 1 and 2 we get
`2I=int_(0)^(pi//2) (sqrt(tanx)+sqrt(cotx))dx`
`=sqrt(2)int_(0)^(pi//2) (sinx+cosx)/(sqrt(sin2x)) dx`
`=sqrt(2) int_(0)^(pi//2) (sinx+cosx)/(sqrt(1-(sinx-cosx)^(2)))dx`
`=sqrt(2)int_(-1)^(1)(dt)/(sqrt(1-t^(2)))` (Putting `sinx-cosx=1`)
`=2sqrt(2)int_(0)^(1)(dt)/(sqrt(1-t^(2)))`
`=2sqrt(2)pi`
or `I=(pi)/(sqrt(2))`