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int 1/(1+cosx) dx=...

`int 1/(1+cosx) dx=`

A

`-cot x + cosec x + C`

B

`cot x - cosec x + C`

C

`cot x + cosec x + C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`I = int{(1)/((1+cosx)) xx ((1-cosx))/((1-cosx))} dx = int((1-cosx))/(sin^(2)x) dx`
`= {(-1)/(sin^(2)x) - (cosx)/(sin^(2)x)} dx = int(cosec^(2)x - cosec x cot x ) dx`.
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Knowledge Check

  • int(x)/(1-cosx)dx=

    A
    `xcot((x)/(2))+2log|sin((x)/(2))|+c`
    B
    `xcot((x)/(2))-2log|sin((x)/(2))|+c`
    C
    `-xcot((x)/(2))+2log|sin((x)/(2))|+c`
    D
    `-xcot((x)/(2))-2log|sin((x)/(2))|+c`

  • int1/sqrt(1+cosx)dx=

    A
    `sqrt(2)logabs(sec""x/2+tan""x/2)+c`
    B
    `1/sqrt(2)logabs(sec""x/2+tan""x/2)+c`
    C
    `logabs(sec""x/2+tan""x/2)+c`
    D
    `2logabs(sec""x/2+tan""x/2)+c`

  • int(1)/(3+2cosx)dx=

    A
    `(2)/(5)tan^(-1)[(tan(x//2))/(5)]+c`
    B
    `(5)/(sqrt2)tan^(-1)[(tan(x//2))/(sqrt2)]+c`
    C
    `(sqrt2)/(5)tan^(-1)[(tan(x//2))/(5)]+c`
    D
    `(2)/(sqrt5)tan^(-1)[(tan(x//2))/(sqrt5)]+c`

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    Evaluate: int1/(cosx-sinx)\ dx

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