Final, Fall 2016

We need to evaluate the definite integral of the function $f(x) = \sin x$ over the interval $[0, \pi/2]$. This involves finding the antiderivative of $\sin x$ and evaluating it at the upper and lower limits of integration using the Fundamental Theorem of Calculus, Part 2.


We need to find a function $F(x)$ such that its derivative $F'(x)$ is equal to $f(x) = \sin x$.
Remember the basic derivative rules for trigonometric functions: the derivative of $\cos x$ is $-\sin x$. Therefore, the antiderivative of $\sin x$ is $-\cos x$, because the derivative of $-\cos x$ is $-(-\sin x) = \sin x$.
So, the antiderivative is $F(x) = -\cos x$.


The Fundamental Theorem of Calculus, Part 2, states that if $F$ is an antiderivative of $f$, then:
$${\int }_a^{b}f(x)\hspace{0.17em}dx=F(b)-F(a)$$
\int_{a}^{b} f(x) \, dx = F(b) - F(a)

In our case, $f(x) = \sin x$, $F(x) = -\cos x$, $a = 0$, and $b = \pi/2$.
$$\begin{align*}$$
& \int_{0}^{\pi/2} \sin x \, dx \\
&= [-\cos x]_{0}^{\pi/2} \\
&= (-\cos(\pi/2)) - (-\cos(0))
\end{align*}



Now we need to evaluate $-\cos x$ at the upper limit $x = \pi/2$ and the lower limit $x = 0$.
Recall the values of cosine at these key angles: $\cos(\pi/2) = 0$ and $\cos(0) = 1$.
Substitute these values into the expression:
$$\begin{align*}$$
& (-\cos(\pi/2)) - (-\cos(0)) \\
&= (-0) - (-1) \\
&= 0 + 1 \\
&= 1
\end{align*}


The value of the definite integral is 1.

$$\boxed{{\displaystyle {\int }_0^{\pi \mathrm{/}2}\sin x\hspace{0.17em}dx=1}}$$
\boxed{ \int_{0}^{\pi/2} \sin x \, dx = 1 }

We need to compute the indefinite integral $\int x\sqrt{9+x^2} \, dx$. The integrand involves a function inside a square root, $9+x^2$, and its derivative (up to a constant factor), $2x$, is present as the $x$ term outside the square root. This structure suggests using integration by substitution (u-substitution).


Let $u$ be the inner function:
$$u=9+x^2$$
u = 9+x^2

Now, find the differential $du$ by differentiating $u$ with respect to $x$:
$$\frac{du}{dx}=2x$$
\frac{du}{dx} = 2x

Rearrange to solve for $du$:
$$du=2x\hspace{0.17em}dx$$
du = 2x \, dx

Notice that our integral has $x \, dx$, not $2x \, dx$. We can solve the differential equation for $x \, dx$: $x \, dx = \frac{1}{2} du$. This allows us to substitute perfectly.


Rewrite the original integral using $u$ and $du$.
Substitute $\sqrt{9+x^2} = \sqrt{u} = u^{1/2}$ and $x \, dx = \frac{1}{2} du$:
$$\begin{align*}$$
& \int x\sqrt{9+x^2} \, dx \\
&= \int \sqrt{u} \left( \frac{1}{2} du \right) \\
&= \frac{1}{2} \int u^{1/2} \, du
\end{align*}



Now integrate $u^{1/2}$ using the power rule for integration: $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ (for $n \neq -1$).
Here, $n = 1/2$, so $n+1 = 3/2$.
$$\begin{align*}$$
& \frac{1}{2} \int u^{1/2} \, du \\
&= \frac{1}{2} \left( \frac{u^{1/2 + 1}}{1/2 + 1} \right) + C \\
&= \frac{1}{2} \left( \frac{u^{3/2}}{3/2} \right) + C \\
&= \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C \\
&= \frac{1}{3} u^{3/2} + C
\end{align*}

Don't forget the constant of integration, $+ C$, when computing indefinite integrals!


Replace $u$ with its expression in terms of $x$, which is $u = 9+x^2$.
$$\frac{1}{3}(9+x^2{)}^{3\mathrm{/}2}+C$$
\frac{1}{3} (9+x^2)^{3/2} + C


The indefinite integral is:
$$\boxed{{\displaystyle \int x\sqrt{9+x^2}\hspace{0.17em}dx=\frac{1}{3}(9+x^2{)}^{3\mathrm{/}2}+C}}$$
\boxed{ \int x\sqrt{9+x^2} \, dx = \frac{1}{3} (9+x^2)^{3/2} + C }

We need to compute the definite integral $\int_{0}^{1} x^2 e^x \, dx$. The integrand is a product of a polynomial function ($x^2$) and an exponential function ($e^x$). Standard u-substitution doesn't apply easily. This suggests using integration by parts. The formula is $\int u \, dv = uv - \int v \, du$.


We need to choose $u$ and $dv$. Using the LIPET/LIATE guideline (Polynomial comes before Exponential), we choose:
Let $u = x^2$ and $dv = e^x \, dx$.
Then we find $du$ by differentiating $u$, and $v$ by integrating $dv$:
$du = 2x \, dx$
$v = \int e^x \, dx = e^x$

Now apply the integration by parts formula:
$$\begin{align*}$$
& \int x^2 e^x \, dx \\
&= uv - \int v \, du \\
&= x^2 e^x - \int e^x (2x \, dx) \\
&= x^2 e^x - 2 \int x e^x \, dx
\end{align*}

Notice that the new integral $\int x e^x \, dx$ is still a product of a polynomial ($x$) and an exponential ($e^x$). This means we need to apply integration by parts a second time.


Let's evaluate the new integral $\int x e^x \, dx$ using integration by parts again.
Choose $u_1 = x$ and $dv_1 = e^x \, dx$.
Then $du_1 = dx$ and $v_1 = \int e^x \, dx = e^x$.
Apply the formula $\int u_1 \, dv_1 = u_1 v_1 - \int v_1 \, du_1$:
$$\begin{align*}$$
& \int x e^x \, dx \\
&= x e^x - \int e^x \, dx \\
&= x e^x - e^x
\end{align*}



Substitute the result from Step 3 back into the expression from Step 2:
$$\begin{align*}$$
& \int x^2 e^x \, dx \\
&= x^2 e^x - 2 \int x e^x \, dx \\
&= x^2 e^x - 2 (x e^x - e^x) + C \\
&= x^2 e^x - 2x e^x + 2e^x + C \\
&= (x^2 - 2x + 2)e^x + C
\end{align*}

So, the antiderivative is $F(x) = (x^2 - 2x + 2)e^x$.


Now we evaluate the definite integral using the limits $a = 0$ and $b = 1$.
$${\int }_0^1{x}^2{e}^x\hspace{0.17em}dx=F(1)-F(0)$$
\int_{0}^{1} x^2 e^x \, dx = F(1) - F(0)

First, calculate $F(1)$:
$$\begin{align*}$$
F(1) &= (1^2 - 2(1) + 2)e^1 \\
&= (1 - 2 + 2)e \\
&= (1)e = e
\end{align*}

Next, calculate $F(0)$:
$$\begin{align*}$$
F(0) &= (0^2 - 2(0) + 2)e^0 \\
&= (0 - 0 + 2)(1) \\
&= 2
\end{align*}

Remember that $e^0 = 1$. This is a common point where errors can occur.
Finally, compute $F(1) - F(0)$:
$$F(1)-F(0)=e-2$$
F(1) - F(0) = e - 2


The value of the definite integral is $e - 2$.

$$\boxed{{\displaystyle {\int }_0^1{x}^2{e}^x\hspace{0.17em}dx=e-2}}$$
\boxed{ \int_{0}^{1} x^2 e^x \, dx = e - 2 }

We need to compute the derivative of a definite integral where the upper limit of integration is a function of $x$. This requires the Fundamental Theorem of Calculus, Part 1 (FTC Part 1) combined with the Chain Rule.


FTC Part 1 states: $\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x)$.
When the upper limit is a function $g(x)$, the combined formula is:
$$\frac{d}{dx}{\int }_a^{g(x)}f(t)\hspace{0.17em}dt=f(g(x))\cdot g^{\mathrm{\prime }}(x)$$
\frac{d}{dx} \int_{a}^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x)

Think of it like this: FTC Part 1 gives you $f(\text{upper limit})$, and then the Chain Rule requires you to multiply by the derivative of the upper limit.


In our problem $\frac{d}{dx} \left( \int_{1}^{x^2} \cos t \, dt \right)$:
The lower limit is a constant: $a = 1$.
The integrand function is: $f(t) = \cos t$.
The upper limit is a function of $x$: $g(x) = x^2$.


First, find the derivative of the upper limit, $g'(x)$:
$$g^{\mathrm{\prime }}(x)=\frac{d}{dx}(x^2)=2x$$
g'(x) = \frac{d}{dx}(x^2) = 2x

Next, evaluate the integrand $f(t)$ at the upper limit $g(x)$:
$$f(g(x))=f(x^2)=\cos (x^2)$$
f(g(x)) = f(x^2) = \cos(x^2)

Now, substitute these pieces into the combined formula $f(g(x)) \cdot g'(x)$:
$$\frac{d}{dx}({\int }_1^{x^2}\cos t\hspace{0.17em}dt)=f(g(x))\cdot g^{\mathrm{\prime }}(x)=\cos (x^2)\cdot (2x)$$
\frac{d}{dx} \left( \int_{1}^{x^2} \cos t \, dt \right) = f(g(x)) \cdot g'(x) = \cos(x^2) \cdot (2x)

It's conventional to write the polynomial factor first.
$$\cos (x^2)\cdot (2x)=2x\cos (x^2)$$
\cos(x^2) \cdot (2x) = 2x \cos(x^2)

Note that the lower limit $a=1$ being a constant doesn't affect the result when differentiating with respect to x. If the lower limit were also a function of x, the formula would be $\frac{d}{dx} \int_{h(x)}^{g(x)} f(t) \, dt = f(g(x))g'(x) - f(h(x))h'(x)$.

The derivative is $2x \cos(x^2)$.

$$\boxed{{\displaystyle \frac{d}{dx}({\int }_1^{x^2}\cos t\hspace{0.17em}dt)=2x\cos (x^2)}}$$
\boxed{ \frac{d}{dx} \left( \int_{1}^{x^2} \cos t \, dt \right) = 2x \cos(x^2) }

The integrand is $\frac{\sin 2x}{1 + \cos^2 x}$. We can simplify the numerator using the double angle identity for sine:
Recall the identity: $\sin 2x = 2 \sin x \cos x$.
Substituting this into the integral expression gives:
$$\frac{2\sin x\cos x}{1+{\cos }^{2}x}$$
\frac{2 \sin x \cos x}{1 + \cos^2 x}

So the integral becomes:
$$\int \frac{2\sin x\cos x}{1+{\cos }^{2}x}\hspace{0.17em}dx$$
\int \frac{2 \sin x \cos x}{1 + \cos^2 x} \, dx



Look at the modified integrand $\frac{2 \sin x \cos x}{1 + \cos^2 x}$. Let's check the derivative of the denominator $1 + \cos^2 x$ using the chain rule:
$$\begin{align*}$$
& \frac{d}{dx} (1 + \cos^2 x) \\
&= 0 + \frac{d}{dx} ((\cos x)^2) \\
&= 2(\cos x)^1 \cdot \frac{d}{dx}(\cos x) \\
&= 2 \cos x (-\sin x) \\
&= -2 \sin x \cos x
\end{align*}

The derivative of the denominator is $-2 \sin x \cos x$, which is the negative of the numerator ($2 \sin x \cos x$). This confirms that integration by substitution (u-substitution) is the appropriate technique, matching the pattern $\int \frac{-du}{u}$.


Let $u$ be the denominator:
$$u=1+{\cos }^{2}x$$
u = 1 + \cos^2 x

Find the differential $du$:
$$du=-2\sin x\cos x\hspace{0.17em}dx$$
du = -2 \sin x \cos x \, dx



Rewrite the integral in terms of $u$. We have $1 + \cos^2 x = u$. The numerator and differential term is $2 \sin x \cos x \, dx$. Comparing this to $du = -2 \sin x \cos x \, dx$, we see that:
$$2\sin x\cos x\hspace{0.17em}dx=-du$$
2 \sin x \cos x \, dx = -du

Substitute these into the integral:
$$\begin{align*}$$
& \int \frac{2 \sin x \cos x}{1 + \cos^2 x} \, dx \\
&= \int \frac{1}{1 + \cos^2 x} (2 \sin x \cos x \, dx) \\
&= \int \frac{1}{u} (-du) \\
&= -\int \frac{1}{u} \, du
\end{align*}



Integrate $\frac{1}{u}$ with respect to $u$:
Remember the integration rule: $\int \frac{1}{u} \, du = \ln|u| + C$.
Applying this rule:
$$\begin{align*}$$
& -\int \frac{1}{u} \, du \\
&= -(\ln|u|) + C \\
&= -\ln|u| + C
\end{align*}



Replace $u$ with its expression in terms of $x$, which is $u = 1 + \cos^2 x$:
$$-\ln \mathrm{\mid }1+{\cos }^{2}x\mathrm{\mid }+C$$
-\ln|1 + \cos^2 x| + C

Since $\cos^2 x \ge 0$, the term $1 + \cos^2 x$ is always positive ($1 + \cos^2 x \ge 1$). Therefore, the absolute value signs are not strictly necessary.
$$-\ln (1+{\cos }^{2}x)+C$$
-\ln(1 + \cos^2 x) + C


The indefinite integral is:
$$\boxed{{\displaystyle -\ln (1+{\cos }^{2}x)+C}}$$
\boxed{ -\ln(1 + \cos^2 x) + C }

We need to compute the indefinite integral $\int \tan^2 \theta \sec^2 \theta \, d\theta$. The integrand is a product of powers of tangent and secant. We should look for a possible u-substitution.
Recall the derivatives of tangent and secant: $\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta$ and $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$.
Since the derivative of $\tan \theta$ is $\sec^2 \theta$, and we have a $\sec^2 \theta$ factor in the integrand, the substitution $u = \tan \theta$ is the most direct approach.


Let $u = \tan \theta$.
Find the differential $du$:
$$du=\frac{d}{d\theta }(\tan \theta )\hspace{0.17em}d\theta ={\sec }^2\theta \hspace{0.17em}d\theta$$
du = \frac{d}{d\theta}(\tan \theta) \, d\theta = \sec^2 \theta \, d\theta

Now substitute $u$ for $\tan \theta$ and $du$ for $\sec^2 \theta \, d\theta$ in the integral:
$$\begin{align*}$$
& \int \tan^2 \theta \sec^2 \theta \, d\theta \\
&= \int (\tan \theta)^2 (\sec^2 \theta \, d\theta) \\
&= \int u^2 \, du
\end{align*}



The integral is now a simple power function of $u$. Integrate using the power rule for integration: $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$.
Here, $n = 2$, so $n+1 = 3$.
$$\begin{align*}$$
& \int u^2 \, du \\
&= \frac{u^{2+1}}{2+1} + C \\
&= \frac{u^3}{3} + C
\end{align*}



Replace $u$ with its expression in terms of $\theta$, which is $u = \tan \theta$.
$$\frac{(\tan \theta {)}^3}{3}+C=\frac{{\tan }^3\theta }{3}+C$$
\frac{(\tan \theta)^3}{3} + C = \frac{\tan^3 \theta}{3} + C


The indefinite integral is:
$$\boxed{{\displaystyle \int {\tan }^2\theta {\sec }^2\theta \hspace{0.17em}d\theta =\frac{{\tan }^3\theta }{3}+C}}$$
\boxed{ \int \tan^2 \theta \sec^2 \theta \, d\theta = \frac{\tan^3 \theta}{3} + C }

The integral is $\int_{2}^{4} \frac{dx}{x^2 - 2x - 3}$. An integral can be improper if the integrand has an infinite discontinuity within the interval of integration or at its endpoints. This happens when the denominator is zero.
Factor the denominator:
$$x^2-2x-3=(x-3)(x+1)$$
x^2 - 2x - 3 = (x-3)(x+1)

The denominator is zero when $x = 3$ or $x = -1$.


The interval of integration is $[2, 4]$.
The value $x = 3$ lies within this interval ($2 < 3 < 4$).
The value $x = -1$ is outside this interval.
Since there is an infinite discontinuity at $x = 3$, which is inside the interval $[2, 4]$, this is an improper integral of Type 2.


To analyze convergence, we must split the integral into two parts at the point of discontinuity, $x = 3$:
$$\begin{array}{rl}{\displaystyle {\int }_2^4\frac{dx}{(x-3)(x+1)}}& {\displaystyle ={\int }_2^3\frac{dx}{(x-3)(x+1)}}\\ {\displaystyle }& {\displaystyle +{\int }_3^4\frac{dx}{(x-3)(x+1)}}\end{array}$$
\begin{align*}
\int_{2}^{4} \frac{dx}{(x-3)(x+1)} &= \int_{2}^{3} \frac{dx}{(x-3)(x+1)} \\
&\qquad + \int_{3}^{4} \frac{dx}{(x-3)(x+1)}
\end{align*}

The original integral converges if and only if *both* of these new integrals converge. If either one diverges, the original integral diverges. Let's analyze the first part, $\int_{2}^{3} \frac{dx}{(x-3)(x+1)}$.


We can use the Limit Comparison Test (LCT) for improper integrals to determine convergence or divergence without calculating the integral's value. We need to compare our integrand, $f(x) = \frac{1}{(x-3)(x+1)}$, to a simpler function $g(x)$ whose convergence behavior near $x=3$ is known.
As $x \to 3$, the factor $(x+1)$ approaches $3+1 = 4$. So, the integrand behaves like $\frac{1}{(x-3)(4)}$. This suggests comparing with:
$$g(x)=\frac{1}{x-3}$$
g(x) = \frac{1}{x-3}

The key idea for LCT is to choose a comparison function $g(x)$ that captures the "most problematic" part of $f(x)$ near the discontinuity. Here, the $(x-3)$ term causes the division by zero.


Calculate the limit of the ratio $f(x) / g(x)$ as $x$ approaches the discontinuity $x=3$ (from the left, for the first integral):
$$\begin{align*}$$
L &= \lim_{x \to 3^{-}} \frac{f(x)}{g(x)} \\
&= \lim_{x \to 3^{-}} \frac{1 / ((x-3)(x+1))}{1 / (x-3)} \\
&= \lim_{x \to 3^{-}} \frac{1}{(x-3)(x+1)} \times \frac{(x-3)}{1} \\
&= \lim_{x \to 3^{-}} \frac{1}{x+1} \\
&= \frac{1}{3+1} = \frac{1}{4}
\end{align*}

Since the limit $L = 1/4$ is a finite positive number ($0 < L < \infty$), the LCT states that $\int_{2}^{3} f(x) \, dx$ and $\int_{2}^{3} g(x) \, dx$ have the same convergence behavior (they either both converge or both diverge).


Now we need to determine if the comparison integral $\int_{2}^{3} g(x) \, dx = \int_{2}^{3} \frac{dx}{x-3}$ converges or diverges.
This is a p-integral with a discontinuity at the upper endpoint. Let $y = x-3$. As $x \to 3^{-}$, $y \to 0^{-}$. The integral behaves like $\int_{a}^{0} \frac{dy}{y^p}$.
In our case, $\int_{2}^{3} \frac{dx}{(x-3)^1}$, the power is $p = 1$.
An improper integral of the form $\int_{a}^{c} \frac{dx}{(c-x)^p}$ or $\int_{c}^{b} \frac{dx}{(x-c)^p}$ converges if $p < 1$ and diverges if $p \ge 1$.
Since $p = 1$, the integral $\int_{2}^{3} \frac{dx}{x-3}$ diverges.


Based on the LCT:
1. We compared $\int_{2}^{3} \frac{dx}{(x-3)(x+1)}$ with $\int_{2}^{3} \frac{dx}{x-3}$.
2. The limit of the ratio of their integrands as $x \to 3^{-}$ was $L = 1/4$ (finite and positive).
3. The comparison integral $\int_{2}^{3} \frac{dx}{x-3}$ diverges (since $p=1$).
Therefore, the integral $\int_{2}^{3} \frac{dx}{(x-3)(x+1)}$ also diverges.

Since one of the parts ($\int_{2}^{3} \dots$) diverges, the original integral $\int_{2}^{4} \frac{dx}{x^2 - 2x - 3}$ must also diverge. There is no need to check the convergence of the second part ($\int_{3}^{4} \dots$).

$$\boxed{{\displaystyle \text{The integral }{\int }_2^4\frac{dx}{x^2-2x-3}\text{ diverges.}}}$$
\boxed{ \text{The integral } \int_{2}^{4} \frac{dx}{x^2 - 2x - 3} \text{ diverges.} }



Beyond the steps to solve this specific problem, what's the bigger picture?

1. Spotting the Trouble: When integrating a fraction, always check if the denominator can be zero within your integration bounds. If it can, like $x^2 - 2x - 3 = (x-3)(x+1)$ being zero at $x=3$ in the interval $[2, 4]$, you've found a vertical asymptote. This means the function value shoots off to infinity, and you're dealing with an improper integral.

2. Does Infinite Height Mean Infinite Area? Not always! A function can go to infinity, but the area underneath it might still be a finite number. Think about $\int_{0}^{1} \frac{1}{\sqrt{x}} dx$: $1/\sqrt{x}$ goes to infinity at $x=0$, but the integral converges (area is finite).

3. It's All About Speed (the Power 'p'): What really matters is *how fast* the function runs towards infinity near the asymptote. This speed is determined by the power $p$ on the factor causing the problem. Near our asymptote $x=3$, our function $\frac{1}{(x-3)(x+1)}$ behaves like $\frac{1}{4(x-3)^1}$. The critical factor is $(x-3)$ raised to the power $p=1$.

4. The p-Test Threshold: There's a crucial rule for integrals near an asymptote $x=c$ that behave like $\frac{1}{|x-c|^p}$:
* If $p \ge 1$, the function grows too fast; the area is infinite (integral diverges).
* If $p < 1$, the function grows slowly enough; the area is finite (integral converges).
Our example had $p=1$, which is right on the edge, and it diverges. This is the same reason the harmonic series $1 + 1/2 + 1/3 + \dots$ diverges. If the factor had been $\sqrt{x-3}$ (so $p=1/2$), the integral would have converged. If it had been $(x-3)^2$ (so $p=2$), it would also diverge.

5. Why Factoring and LCT are Useful: Factoring the denominator helps you find the asymptotes and identify the crucial power $p$. The Limit Comparison Test (LCT) is a tool that formalizes the idea of "behaves like". It lets you compare your complicated function to the simpler $\frac{1}{|x-c|^p}$ form near the asymptote to determine convergence or divergence without actually finding the antiderivative.

The integral is $\int_{0}^{\infty} (1 + \sin(e^x)) e^{-2x} \, dx$. Since the upper limit of integration is $\infty$, this is an improper integral of Type 1.


Let $f(x) = (1 + \sin(e^x)) e^{-2x}$. We need to understand its behavior, especially as $x \to \infty$.
The term $\sin(e^x)$ oscillates as $x$ increases, since $e^x$ goes to infinity. However, we know that the sine function is always bounded:
$$-1\le \sin (\theta )\le 1\text{for any }\theta$$
-1 \le \sin(\theta) \le 1 \quad \text{for any } \theta

So, for $\theta = e^x$, we have $-1 \le \sin(e^x) \le 1$.
This allows us to bound the term $(1 + \sin(e^x))$:
$$1+(-1)\le 1+\sin (e^x)\le 1+(1)$$
1 + (-1) \le 1 + \sin(e^x) \le 1 + (1)

$$0\le 1+\sin (e^x)\le 2$$
0 \le 1 + \sin(e^x) \le 2

The exponential term $e^{-2x}$ is always positive for real $x$.
Therefore, we can bound the entire integrand $f(x)$:
Multiply the inequality by $e^{-2x}$:
$$0\cdot e^{-2x}\le (1+\sin (e^x))e^{-2x}\le 2\cdot e^{-2x}$$
0 \cdot e^{-2x} \le (1 + \sin(e^x)) e^{-2x} \le 2 \cdot e^{-2x}

$$0\le f(x)\le 2e^{-2x}$$
0 \le f(x) \le 2e^{-2x}

This inequality holds for all $x \ge 0$.


We have found that our integrand $f(x)$ is non-negative and is less than or equal to $g(x) = 2e^{-2x}$. This setup is ideal for the Direct Comparison Test for improper integrals.
The Direct Comparison Test states: If $0 \le f(x) \le g(x)$ for $x \ge a$, then if $\int_{a}^{\infty} g(x) \, dx$ converges, $\int_{a}^{\infty} f(x) \, dx$ must also converge.
Our comparison function is $g(x) = 2e^{-2x}$.


We need to determine if the integral of the comparison function, $\int_{0}^{\infty} g(x) \, dx = \int_{0}^{\infty} 2e^{-2x} \, dx$, converges or diverges.
This is a standard exponential integral. We can evaluate it directly:
$$\begin{align*}$$
& \int_{0}^{\infty} 2e^{-2x} \, dx \\
&= \lim_{b \to \infty} \int_{0}^{b} 2e^{-2x} \, dx \\
&= \lim_{b \to \infty} \left[ 2 \frac{e^{-2x}}{-2} \right]_{0}^{b} \\
&= \lim_{b \to \infty} \left[ -e^{-2x} \right]_{0}^{b} \\
&= \lim_{b \to \infty} (-e^{-2b}) - (-e^{-2(0)}) \\
&= \lim_{b \to \infty} (-e^{-2b}) + e^{0}
\end{align*}

As $b \to \infty$, $-2b \to -\infty$, so $e^{-2b} \to 0$. Also, $e^0 = 1$.
$$=(0)+1=1$$
= (0) + 1 = 1

Since the limit is a finite number (1), the integral $\int_{0}^{\infty} 2e^{-2x} \, dx$ converges.
Alternatively, you might know that $\int_{0}^{\infty} e^{-kx} dx$ converges for any constant $k > 0$. Here $k=2$, so $\int_{0}^{\infty} e^{-2x} dx$ converges, and multiplying by 2 doesn't change convergence.


We established:
1. $0 \le (1 + \sin(e^x)) e^{-2x} \le 2e^{-2x}$ for all $x \ge 0$.
2. $\int_{0}^{\infty} 2e^{-2x} \, dx$ converges.

According to the Direct Comparison Test, since our function's integral is bounded above by a convergent integral, our integral must also converge.


The integral $\int_{0}^{\infty} (1 + \sin(e^x)) e^{-2x} \, dx$ converges by the Direct Comparison Test, comparing it with the convergent integral $\int_{0}^{\infty} 2e^{-2x} \, dx$.

$$\boxed{{\displaystyle \text{The integral converges.}}}$$
\boxed{ \text{The integral } \text{ converges.} }

Let the terms of the series be $a_n = \frac{n^2 (1+\cos^2(n))}{n^5+3}$.
We need to determine if the infinite series $\sum_{n=1}^{\infty} a_n$ converges or diverges.
First, consider the terms $a_n$. Since $n \ge 1$, $n^2 > 0$. The term $\cos^2(n) \ge 0$, so $1+\cos^2(n) > 0$. The denominator $n^5+3 > 0$. Therefore, all terms $a_n$ are positive ($a_n > 0$). This allows us to use comparison tests.


The term $(1+\cos^2(n))$ oscillates, which can make tests like the Limit Comparison Test difficult if the limit doesn't exist. However, we know the cosine function is bounded.
Recall that $-1 \le \cos(n) \le 1$. Squaring gives $0 \le \cos^2(n) \le 1$.
Therefore, the term $(1+\cos^2(n))$ is bounded:
$$1+0\le 1+{\cos }^2(n)\le 1+1$$
1 + 0 \le 1 + \cos^2(n) \le 1 + 1

$$1\le 1+{\cos }^2(n)\le 2$$
1 \le 1 + \cos^2(n) \le 2

We can use the upper bound to establish an inequality for $a_n$:
$$\begin{align*}$$
a_n &= \frac{n^2 (1+\cos^2(n))}{n^5+3} \\
&\le \frac{n^2 (2)}{n^5+3} \\
&= \frac{2n^2}{n^5+3}
\end{align*}

Now, we can simplify this further by noting that for $n \ge 1$, $n^5+3 > n^5$. Therefore, $\frac{1}{n^5+3} < \frac{1}{n^5}$.
$$\begin{align*}$$
a_n &\le \frac{2n^2}{n^5+3} \\
&< \frac{2n^2}{n^5} \\
&= \frac{2}{n^3}
\end{align*}

So, we have established that $0 < a_n < \frac{2}{n^3}$ for all $n \ge 1$.
This suggests comparing our series to the series $\sum_{n=1}^{\infty} \frac{2}{n^3}$.


Let $b_n = \frac{2}{n^3}$. The comparison series is $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{2}{n^3}$.
This can be written as $2 \sum_{n=1}^{\infty} \frac{1}{n^3}$.
The series $\sum_{n=1}^{\infty} \frac{1}{n^3}$ is a p-series with $p = 3$.
Recall that a p-series $\sum \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$.
Since $p = 3 > 1$, the p-series $\sum_{n=1}^{\infty} \frac{1}{n^3}$ converges.
Therefore, the comparison series $\sum_{n=1}^{\infty} b_n = 2 \sum_{n=1}^{\infty} \frac{1}{n^3}$ also converges (multiplying by a constant doesn't affect convergence).


We have shown that:
1. $0 < a_n < b_n$ for all $n \ge 1$, where $a_n = \frac{n^2 (1+\cos^2(n))}{n^5+3}$ and $b_n = \frac{2}{n^3}$.
2. The series $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{2}{n^3}$ converges.

The Direct Comparison Test (DCT) states that if $0 \le a_n \le b_n$ for all sufficiently large $n$, and $\sum b_n$ converges, then $\sum a_n$ must also converge.
Think of it like this: If the terms of your series ($a_n$) are always positive but smaller than the terms of a series that adds up to a finite number ($\sum b_n$), then your series must also add up to a finite number.
Since both conditions of the DCT are met, our series $\sum a_n$ converges.


The series $\sum_{n=1}^{\infty} \frac{n^2 (1+\cos^2(n))}{n^5+3}$ converges based on the Direct Comparison Test with the convergent p-series $\sum_{n=1}^{\infty} \frac{2}{n^3}$.