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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

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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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