91Ó°ÊÓ

First, recall the integration by parts formula: \(\int u dv = uv - \int v du\). Choose \(u\) and \(dv\) as follows: - Let \(u = e^{-x} \Rightarrow du = -e^{-x} dx\) - Let \(dv = \cos^2 x \, dx \Rightarrow v = \int \cos^2 x \,dx\)

To find \(v\), we need to integrate \(\cos ^2 x\) with respect to \(x\). We can use the double-angle formula: \(\cos 2x = 2\cos^2 x - 1\) to express \(\cos^2 x\) in terms of \(\cos 2x\): $$\cos^2 x = \frac{1+\cos 2x}{2}$$ Now we are ready to integrate \(\cos^2 x\): $$v = \int \cos^2 x \, dx = \int \frac{1+\cos 2x}{2} \, dx = \frac{1}{2} \int (1+\cos 2x) \, dx$$ Evaluate the integral: $$v = \frac{1}{2} \left[ \int 1\, dx + \int \cos 2x\, dx \right] = \frac{1}{2} \left[ x + \frac{1}{2} \sin 2x \right] + C$$

Now, we apply the integration by parts formula on our original integral: $$\int e^{-x} \cos^{2} x d x = u v \Big|_{0}^{\infty} - \int v du$$ First, let's evaluate \(uv\Big|_{0}^{\infty}\): $$\lim_{x\to\infty} (e^{-x}\left(\frac{1}{2}x+\frac{1}{2}\sin 2x\right)) - (e^{-0}\left(\frac{1}{2}\cdot 0 + \frac{1}{2}\sin 0\right)) = 0 - 0 =0$$ Since the first part evaluates to 0, let's simplify the second integral: $$-\int v du = -\int \left(\frac{1}{2} x +\frac{1}{2} \sin 2x \right) (-e^{-x}) dx = \int \left(\frac{1}{2} x +\frac{1}{2} \sin 2x \right) e^{-x} dx$$

Now we need to apply integration by parts again. Choose \(u\) and \(dv\) as follows: - Let \(u = x e^{-x} \Rightarrow du = (1-x)e^{-x} dx\) - Let \(dv = \frac{1}{2} \sin 2x \, dx \Rightarrow v = -\frac{1}{4} \cos 2x\) Apply the integration by parts formula: $$ \int x e^{-x} \sin 2x\, dx = -\frac{1}{4}x e^{-x} \cos 2x - \int -\frac{1}{4} \cos 2x (1-x)e^{-x} dx$$ The remaining integral can be evaluated directly: $$-\frac{1}{4} \int (1-x)e^{-x} \cos 2x dx = -\frac{1}{4} \int e^{-x} \cos 2x dx + \frac{1}{4} \int x e^{-x} \cos 2x dx$$ Combine the integral terms together: $$\int x e^{-x} \sin 2x\, dx = -\frac{1}{4} x e^{-x} \cos 2x + \frac{1}{4} \int e^{-x} \cos 2x dx - \frac{1}{4} \int x e^{-x} \cos 2x dx$$ Adding both the integrals form, we get: $$\frac{5}{4} \int x e^{-x} \sin 2x dx = -\frac{1}{4} x e^{-x} \cos 2x + \frac{1}{4} \int e^{-x} \cos 2x dx$$ Calculate the remaining integral: $$\int e^{-x} \cos 2x dx = \int e^{-x} \sin 2x dx + \frac{1}{2} x e^{-x} \cos 2x$$

Finally, combine all the results to form the final answer: $$\int e^{-x} \cos^{2} x d x = \frac{1}{2}x+\frac{1}{2}\sin 2x+C_1$$