Advertisements
Advertisements
Question
Evaluate : `int _0^(pi/4) 1/(1 + "x"^2) "dx"`
Sum
Advertisements
Solution
Let I = `int _0^(pi/4) 1/(1 + "x"^2) "dx"`
`= ["tan"^(-1) "x"]_0^(pi/4)`
`= "tan"^(-1) (pi/4) - "tan"^(-1) (0)`
= 1 - 0
= 1
shaalaa.com
Is there an error in this question or solution?
