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int(0)^(pi//2)x cot x dx=(pi)/(2)(log2)...

`int_(0)^(pi//2)x cot x dx=(pi)/(2)(log2)`

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To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} x \cot x \, dx \), we will use integration by parts. ### Step 1: Set up integration by parts We choose: - \( u = x \) (which implies \( du = dx \)) - \( dv = \cot x \, dx \) (which implies \( v = \log(\sin x) \)) ### Step 2: Apply integration by parts formula ...
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Prove that: int_(0)^(pi//2) log (sin x) dx =int_(0)^(pi//2) log (cos x) dx =(-pi)/(2) log 2

Prove that int_(0)^(pi//2)log (sinx)dx=int_(0)^(pi//2) log (cosx)dx=-(pi)/(2) log 2 .

Knowledge Check

  • int_(0)^(pi//2) x cot dx =

    A
    `(pi)/(2) log 2`
    B
    `pi log 2`
    C
    `2pi log 2`
    D
    `-(pi)/(2) log 2`

  • If int_(0)^(pi//2) log cos x dx =(pi)/(2)log ((1)/(2)), then int_(0)^(pi//2) log sec x dx =

    A
    `(pi)/(2)log ""((1)/(2))`
    B
    `1-(pi)/(2)log ""((1)/(2))`
    C
    `1+(pi)/(2) log ""((1)/(2))`
    D
    `(pi)/(2) log 2`

  • int_(0)^(pi//2)log (sec x) dx=

    A
    `(-pi)/2log2`
    B
    `pi/2 log 2`
    C
    `(-pi)/4 log 2`
    D
    `pi/4log2`

    • Similar Questions

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    Show that int_(0)^((pi)/(2))log(sin2x)dx=-(pi)/(2)(log2)

    int_(0)^(pi//2) log (tan x ) dx=

    If int_(0)^(pi//2) log(cosx) dx=pi/2 log (1/2), then int_(0) ^(pi//2) log (sec x ) dx =

    If int_(0)^((pi)/2)log(cosx)dx=-(pi)/2log2 , then int_(0)^((pi)/2)log(cosecx)dx=

    int_(pi//4)^(pi//2)cot^(2)x dx =

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