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int (sec^(2) x)/(" cosec "^(2)x) dx...

`int (sec^(2) x)/(" cosec "^(2)x) dx`

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To solve the integral \(\int \frac{\sec^2 x}{\csc^2 x} \, dx\), we can follow these steps: ### Step 1: Rewrite the integral We know that: \[ \sec^2 x = \frac{1}{\cos^2 x} \quad \text{and} \quad \csc^2 x = \frac{1}{\sin^2 x} \] Thus, we can rewrite the integral as: ...
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Knowledge Check

  • int (sec^(8)x)/("cosec"x)dx=

    A
    `(sec^(8)x)/(8) +c`
    B
    `(sec^(7)x)/(7)+c`
    C
    `(sec^(6)x)/(6)+c`
    D
    `(sec^(9)x)/(9)+c`

  • int(sec^(8)x)/("cosec x")dx=

    A
    `(sec^(8)x)/(8)+c`
    B
    `(sec^(7)x)/(7)+c`
    C
    `(sec^(6)x)/(6)+c`
    D
    `(sec^(9)x)/(9)+c`

  • The integral int sec^(2//3) "x cosec"^(4//3)"x dx" is equal to (here C is a constant of integration)

    A
    `3tan^(-1//3)x+C`
    B
    `-3tan^(-1//3)x+C`
    C
    `-3cot^(-1//3)x+C`
    D
    `-(3)/(4)tan^(-4//3)x+C`

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    int sec^(2//3) x cosec^(4//3) x dx is equal to

    int ("cosec"^(2)x)/((1-cot^(2)x))dx=?

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