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Evaluate: int(sinx)/(1+sinx)\ dx...

Evaluate: `int(sinx)/(1+sinx)\ dx`

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To evaluate the integral \( \int \frac{\sin x}{1 + \sin x} \, dx \), we can follow these steps: ### Step 1: Multiply by the Conjugate We start by multiplying the numerator and denominator by the conjugate of the denominator, which is \( 1 - \sin x \): \[ \int \frac{\sin x}{1 + \sin x} \, dx = \int \frac{\sin x (1 - \sin x)}{(1 + \sin x)(1 - \sin x)} \, dx \] ...
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Knowledge Check

  • int(dx)/(1-2sinx)=

    A
    `(1)/(2sqrt(3))log|(tan((x)/(2))-2+sqrt(3))/(tan((x)/(2))-2-sqrt(3))|+c`
    B
    `(1)/(2sqrt(3))log|(tan((x)/(2))-2-sqrt(3))/(tan((x)/(2))-2+sqrt(3))|+c`
    C
    `(1)/(sqrt(3))log|(tan((x)/(2))-2-sqrt(3))/(tan((x)/(2))-2+sqrt(3))|+c`
    D
    `(1)/(sqrt(3))log|(tan((x)/(2))-2+sqrt(3))/(tan((x)/(2))-2-sqrt(3))|+c`

  • The value of int(1+sinx)/(1-sinx)dx

    A
    `2 tan ((x)/(2) + (pi)/(4)) + C`
    B
    `2 tan ((x)/(2) + (pi)/(4)) + x +C`
    C
    `2 tan ((x)/(2) + (4)/(pi)) - x + C`
    D
    `2 tan^(2) ((x)/(pi)/(4)) - X + C`

  • int(sinx)/(sinx-cosx)dx=

    A
    `(1)/(2).log(sinx-cosx)+c`
    B
    `(1)/(2).[log(sinx-cosx)+x]+c`
    C
    `(1)/(2).log(cosx-sinx)+x+c`
    D
    `(1)/(2).[log(cosx-sinx)+x]+c`

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