91Ó°ÊÓ

To perform the trigonometric substitution, we should pick a substitution that helps simplify the expression in the denominator. Since we have \(1 + y^2\), a suitable substitution would be to let $$y = \tan x.$$

We'll need the differential term \(dy\) in terms of \(x\). Differentiate \(y\) with respect to \(x\): $$\frac{dy}{dx} = \frac{d}{dx} (\tan x),$$ so $$dy = \sec^2 x \, dx.$$

Substitute \(y = \tan x\) and \(dy = \sec^2 x \, dx\) into the given integral. We get $$\int \frac{(\tan^4 x)}{1 + \tan^2 x} \sec^2 x \, dx.$$

Now, we can use the identity \(\sec^2 x = 1 + \tan^2 x\) to simplify the integral expression further. We get $$\int \frac{(\tan^4 x)}{\sec^2 x} \sec^2 x \, dx = \int \tan^4 x \, dx.$$

To evaluate the integral, we can use integration by parts (or use the reduction formula for the integral of \(\tan^n x\)), but we choose another trick by making use of the double-angle formula for \(\tan\): $$\tan^2(2x) = \frac{2\cdot \tan x}{1-\tan^2 x}.$$ This leads to $$\tan^4 x =\frac{(1-\tan^2 x)^2}{4}\cdot \tan^2(2x).$$ Now the integral becomes $$\frac{1}{4} \int (1-\tan^2 x)^2 \tan^2(2x) \, dx.$$ Next, apply the substitution \(u = \tan x\), \(du = \sec^2 x \,dx\): $$ \frac{1}{4} \int (1-u^2)^2 u^2(2 du) = \frac{1}{2} \int (1-u^2)^2 u^2 du. $$ Use polynomial multiplication and integrate term-by-term: $$\frac{1}{2} \int (1 - 2u^2 + u^4)u^2 du = \frac{1}{2} \int (u^2 - 2u^4 + u^6) du.$$ Now integrate each term: $$\frac{1}{2} \left[\frac{1}{3}u^3 - \frac{2}{5}u^5 + \frac{1}{7}u^7\right] + C = \frac{1}{2} \left[\frac{1}{3}\tan^3 x - \frac{2}{5}\tan^5 x + \frac{1}{7}\tan^7 x\right] + C.$$

Finally, reverse the substitution by expressing the result in terms of \(y\) using \(y = \tan x\): $$\frac{1}{2} \left[\frac{1}{3}y^3 - \frac{2}{5}y^5 + \frac{1}{7}y^7\right] + C.$$ So, the integral evaluates to: $$\int \frac{y^4}{1 + y^2} dy = \frac{1}{2} \left[\frac{1}{3}y^3 - \frac{2}{5}y^5 + \frac{1}{7}y^7\right] + C.$$