Explanation:
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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