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int(1)/(sin^(2)x.cos^(2)x)dx is equal to...

`int(1)/(sin^(2)x.cos^(2)x)dx` is equal to

A

`sinx - cos x +C`

B

`tanx +cot x+C`

C

`cos x +sinx +C`

D

`tanx -cot x +C`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the integral \( \int \frac{1}{\sin^2 x \cos^2 x} \, dx \), we can follow these steps: ### Step 1: Rewrite the Integral We start with the integral: \[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx \] This can be rewritten using the identity \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \): \[ \int \frac{4}{\sin^2(2x)} \, dx \] ### Step 2: Use the Cosecant Function We know that \( \frac{1}{\sin^2(2x)} = \csc^2(2x) \). Therefore, we can rewrite the integral as: \[ 4 \int \csc^2(2x) \, dx \] ### Step 3: Integrate The integral of \( \csc^2(kx) \) is \( -\frac{1}{k} \cot(kx) + C \). Here, \( k = 2 \): \[ 4 \int \csc^2(2x) \, dx = 4 \left(-\frac{1}{2} \cot(2x) + C\right) = -2 \cot(2x) + C \] ### Step 4: Final Result Thus, the final result of the integral is: \[ -2 \cot(2x) + C \]

To solve the integral \( \int \frac{1}{\sin^2 x \cos^2 x} \, dx \), we can follow these steps: ### Step 1: Rewrite the Integral We start with the integral: \[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx \] This can be rewritten using the identity \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \): ...
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Knowledge Check

  • int(sin^(2)x - cos^(2)x)/(sin^(2)x cos^(2)x) dx is equal to

    A
    `tan x + cot x + C`
    B
    `tan x + cosec x + C`
    C
    `-tan x + cot x + C`
    D
    `tan x + sec x + C`
  • int 1/(sin^2x cos^2x )dx is equal to:

    A
    `tanx+cotx+c`
    B
    `tanx-cotx+c`
    C
    `tanx.cotx+c`
    D
    `tanx-cot2x+c`
  • int_0^(pi//2) (sin^(1//2)x)/(sin^(1//2)x+cos^(1//2)x) dx is equal to

    A
    0
    B
    `pi/2`
    C
    `pi/3`
    D
    `pi/4`
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