Question

In Exercises \(47-52,\) use the given trigonometric identity to set up a \(u\) -substitution and then evaluate the indefinite integral. $$\int\left(\cos ^{4} x-\sin ^{4} x\right) d x, \quad \cos 2 x=\cos ^{2} x-\sin ^{2} x$$

Short answer

The integral of \( \int\left(\cos^{4} x-\sin^{4} x\right) d x \) is \( x/2 + (1/8)\sin(4x) + C \)

Step-by-step solution

Applying the Identity

From the given identity, one knows that \(\cos^{2}x - \sin^{2}x = \cos 2x\). Square that equation, we get \((\cos^{2}x - \sin^{2}x) ^{2} = \cos^{4}x - 2\cos^{2}x\sin^{2}x + \sin^{4}x = \cos^{2}(2x)\) which implies that \(\cos^{4}x - \sin^{4}x = \cos^{4}x - 2\cos^{2}x\sin^{2}x + \sin^{4}x = \cos^{2}(2x)\) The given integral can now be rewritten as \(\int\cos^{2}(2x)dx\)

Setting up u-substitution

To facilitate integration, a substitution can be made. Set \(u=2x\). Therefore, \(du=2dx\) and \(dx=du/2\). Substituting these values into the integral gives \((1/2)\int\cos^{2}(u)du\)

Evaluating the Integral

The integral of \(\cos^{2}(u)\) can be evaluated using a power-reducing identity that goes as follows: \(\cos^{2}u = {(1 + \cos(2u))}/{2}\). Substituting this into the integral gives \((1/2)\int{(1 + \cos(2u))}/{2}du = (1/4)\int du + (1/4) \int\cos(2u) du = (u/4) + (1/4)(\int\cos(2u) du)\). The integral of \(\cos(2u)\) is \((1/2)\sin(2u)\), hence \((1/4)\int\cos(2u) du = (1/4)(1/2)\sin(2u) = (1/8)\sin(2u)\). Therefore, our integral becomes \((u/4) + (1/8)\sin(2u) + C\)

Unsubstitute u

Substitute back \(u = 2x\) to get the final answer: \(x/2 + (1/8)\sin(2*(2x)) + C = x/2 + (1/8)\sin(4x) + C\)