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Consider the integral I=int(0)^(2pi)(dx)...

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)`
Making the substitution `"tan"1/2x=t`, we have
`I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0`
The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.

Text Solution

Verified by Experts

The correct Answer is:
2
Here the mistake lies in the substitution `"tan"1/2x=t`, because `"tan"1/2x` is discontinuous at `x=pi` which is a point in the interval `[0,2pi]`.
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Consider the integral I = int_(0)^(2pi) (dx)/( 5 - 2 cos x) . Making the substitution tan (x)/(2) = t we have int_(0)^(2pi) (dx)/(5 - 2 cos x) = int_(0)^(0) (2 dt)/((1+ t^(2))(5 - 2 (1-t^(2))/(1+ t^(2))))=0 The result is obviously wrong, since the integrand is positive, and , consequently, the integral of this function cannot be equal to zero . Find the mistake.

Consider the integral int_(0)^(2 pi)(dx)/(5-2cos x) making the substitution tan((x)/(2))=t, we have I=int_(0)^(2 pi)(dx)/(5-2cos x)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))/(1+t^(2))])=0 The result functive and consequently the integral of this functive and consequently the integral of this functake.

Knowledge Check

  • int_(0)^(pi//2)(dx)/(2+cosx)=

    A
    `1/(sqrt(3))tan^(-1)(1/(sqrt(3)))`
    B
    `sqrt(3)tan^(-1)(sqrt(3))`
    C
    `2/(sqrt(3))tan^(-1)(1/(sqrt(3)))`
    D
    `2sqrt(3)tan^(-1)(sqrt(3))`

  • The value of I=int_(0)^(pi//2) (1)/(1+cosx)dx is

    A
    `(pi)/(4)`
    B
    `(pi)/(2)`
    C
    1
    D
    `pi`

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