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Find the integral of (cos^(2)x-sin^(2)x)/(7cos^(2)xsin^(2)x)

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Explanation:

Here's the solution to your question:

We can start by using the trigonometric identity:

cos^2(x) - sin^2(x) = cos(2x)

So the integral becomes:

∫[cos^2(x) - sin^2(x)]/[7cos^2(x)sin^2(x)] dx = ∫[cos(2x)]/[7cos^2(x)sin^2(x)] dx

Next, we can use the substitution u = sin(x), du/dx = cos(x)dx

This gives:

dx = du/cos(x)

And the integral becomes:

∫[cos(2x)]/[7cos^2(x)sin^2(x)] dx = ∫[cos(2x)]/[7u^2(1-u^2)] du/cos(x)

Using the trigonometric identity:

cos(2x) = 2cos^2(x) - 1

The integral further becomes:

∫(2cos^2(x) - 1)dx/[7u^2(1-u^2)] = ∫(2u^2/cos^2(x) - 1/cos^2(x)) du/[7u^2(1-u^2)]

Splitting the integrals gives:

∫(2u^2/cos^2(x))/[7u^2(1-u^2)] du - ∫(1/cos^2(x))/[7u^2(1-u^2)] du

Simplifying each integral gives:

∫(2/cos^2(x))/[7(1-u^2)] du - ∫(1/cos^4(x))/[7u^2(1-u^2)] du

Now we use the trigonometric identity:

cos^2(x) = 1 - sin^2(x)

Substituting back for cos(x) gives:

∫(2/[7(1-u^2)sin^2(x)]) du - ∫(1/[7u^2(1-u^2)(1-sin^2(x))]) du

Using partial fraction decomposition and simplifying gives:

∫(1/7)(1/[u(1+u)(1-u)])[(1+u)/(1-u)] du - ∫(1/7)(1/[u(1+u)])(1/[1-sin^2(x)]) du

The integral further

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