Advertisement Remove all ads

Evaluate :∫π/3 π/6 dx/(1+√cotx) - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`

Advertisement Remove all ads

Solution

Let `I = ∫_(π/3)^(π/6)1/(1 +sqrt(cotx)dx`

`=int_(pi/6)^(pi/3)1/(1+(sqrt(cosx)/sqrtsinx))dx`

`=int_(pi/6)^(pi/3)1/(sqrtsinx+(sqrt(sinx)+sqrtcosx))dx...(1)`

`now,I=int_(pi/6)^(pi/3)(sqrt(sin(pi/2-x)))/(sqrt(sin(pi/2-x))+sqrt(cos(pi/2-x)))dx   .................(int_a^bf(X)dx=int_a^bf(b+a-x)dx)`

`=int_(pi/6)^(pi/3)(sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`

Adding (1) and (2), we get

`2I=int_(pi/6)^(pi/3)(sqrt(sinx)+sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`

`2I=int_(pi/6)^(pi/3)dx`

`2I=[x]_(pi/6)^(pi/3)`

`2I=pi/3-pi/6`

`2I=pi/6`

`I=pi/12`

Concept: Integration Using Trigonometric Identities
  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×