Advertisements
Advertisements
Question
Evaluate :`int_0^(pi/2)dx/(1+cotx)`
Solution
Let `I=int_0^(pi/2)dx/(1+cotx)`
` =int_0^(pi/2)dx/(1+(cosx)/(sinx))`
`=int_0^(pi/2)dx/((sinx+cosx)/sinx)`
`=int_0^(pi/2)sinx/(sinx+cosx)dx..................(i)`
We know
`int_0^af(x)dx=int_0^af(a-x)dx`
`I=int_0^(pi/2)(sin(pi/2-x))/(sin(pi/2-x)+cos(pi/2-x))`
`I=int_0^(pi/2)cosx/(cosx+sinx)dx.................(ii)`
Adding (i) and (ii)
`I+I=int_0^(pi/2)sinx/(sinx+cosx)dx+int_0^(pi/2)cosx/(cosx+sinx)dx`
`2I=int_0^(pi/2)(sinx+cosx)/(sinx+cosx)dx`
`2I=int_0^(pi/2)1dx=[x]_0^(pi/2)`
`I=1/2[x]_0^(pi/2)=1/2[pi/2-0]`
`I=pi/4`
APPEARS IN
RELATED QUESTIONS
evaluate : `int xcosxdx`
Evaluate : `int (2"x" + 1)/(("x" + 1)("x"-2)) "dx"`
If the function f is continuous at x = 2 and x = 4 then find the values of a and b.
Where f(x) = x2 + ax + b, x < 2
= 3x + 2, 2 ≤ x ≤ 4
= 2ax + 5b, 4 < x
Evaluate ; `int x^2/(x^6 - 4x^3 + 13)` dx
`int (2x + 1)/((x + 1) ( x - 2))` dx
Evaluate : `int x cosx dx`
Evaluate : `∫((1 + logx))/(x(2 + logx)(3 + logx)`dx