5.1 Antiderivatives

In this chapter we focus on The Fundamental Theorem of Calculus (TFTC). Central for its understanding is the concept of an antiderivative/indefinite integral of a function. And, central to understand the general form of an antiderivative of a function on an interval are two consequences of the Mean Value Theorem(MVT). Actually, we will also use the MVT to prove part 2 of TFTC.

5.1 Antiderivatives

5.1.1 Two Consequences of the Mean Value Theorem

Theorem 1.

1. Let \(f\) be a differentiable function over an open interval \(I\). If \(f'(x) = 0\) for all \(x \in I\), then \(f \) is constant on \(I\).
2. Let \(f\) and \(g\) be differentiable functions on an open interval \(I\). If \(f'(x) = g'(x)\) for all \(x \in I\), then there exists a constant \(C\in \mathbb{R}\) such that \(f (x) =g(x) + C\) for all \(x \in I\).

Proof.
(1) Let \(x_0\in I\) be fixed, and let \(x\neq x_0\) be a second arbitrary number in \(I\). Then, by the MVT, there exists a point \(x^*\) between \(x_0\) and \(x\) such that
\[f(x) – f(x_0) = f'(x^*)(x-x_0).\]
But, since \(f ‘(x) = 0\) for all \(x \in I\),
\[f(x) – f(x_0) = 0 \Longrightarrow f(x) = f(x_0) \;\; \forall \,x\in I.\]
Thus, \(f\) is constant.

(2) Let \(h= f-g\), then \(h'(x) = f'(x)-g'(x) = 0\) for all \(x\in I\). Then, by part (1), \(h\) is constant, say equal to \(C\). It follows that
\[h(x) = C \;\; \Longrightarrow \;\; f(x)-g(x) = C \;\; \Longrightarrow \;\; f(x) = g(x) + C \;\; \forall\, x\in I.\]

Example 1. Show that
\[\arccos x = -\arcsin x + \frac{\pi}{2} \;\;\; \forall \, x \in [-1,1],\]
and find the value of \(C\).

Solution. Let \(f(x) = \arccos x + \arcsin x\). Then, \(f\) is continuous on \(\displaystyle [-1,1]\), differentiable on \(\displaystyle \left (-1,1\right )\), and \(f'(x) = 0\) (explain why). Thus, by Theorem 1 (1), \(f\) is constant on \((-1,1)\), and, by continuity, on \(\displaystyle \left [-1,1\right ]\). Since \(\displaystyle f(0) = \frac{\pi}{2}+0\), it follows that \(\displaystyle \arccos x +\arcsin x = \frac{\pi}{2} \).

5.1.3 General Form of Antiderivatives

As the name indicates, an antiderivative is the reverse process of computing a derivative. More precisely.

Definition 1.
A function \(F\) is an antiderivative of the function \(f\) if
\[F’ (x) = f (x)\]
for all \(x\) in the domain of \(f\).

Example 1. Verify that the following statements are true:
 

(a) \(F(x) = -\cos x\) is an antiderivative of \(f(x) = \sin x\)
 

(b) \(G(x) = x^3+1\) is an antiderivative of \(g(x) = 3x^2\)
 

(c) \(H(t) = \arcsin 2t\) is an antiderivative of \(\displaystyle h(t) =\frac{2}{\sqrt{1-4t^2}}\).

Solution.
(a) Since \(\displaystyle \frac{d}{dx}\cos x =- \sin x\), then \(\displaystyle \frac{d}{dx}(-\cos x) = -(-\sin x) = \sin x\).

(b) It is immediate to verify that \(G'(x) = g(x)\).

(c) Applying the Chain Rule, we have \(\displaystyle \frac{d}{dt} \arcsin at = \frac{2}{\sqrt{1-4t^2}}\). The result follows.

Observation. Since the derivative of a constant function is zero, adding a constant to the antiderivative of a function produces a new antiderivative. Thus, the functions
\[\sin x, \;\; \sin x + \pi, \;\; \sin x +C \]
\(C\in \mathbb{R}\), constant, are all antiderivatives of \(\cos x\). The following theorem shows that the process of adding a constant to a given antiderivative of a function, essentially renders all antiderivatives.

Theorem 2.
Let \(F\) and \(G\) be antiderivatives of a function \(f\) over an open interval \(I\). Then, there exists a constant \(C\) such that
\[G(x)= F(x) +C\;\;\;\; \forall \, x\in I.\]

Proof. The proof is immediate from Theorem 1(2). Since \(F\) and \(G\) are antiderivatives of \(f\) over \(I\), then \(F'(x)=G'(x)\) for all \(x\in I\). Therefore, there exists a constant \(C\) such that \(G(x) = F(x)+C\) for all \(x \in I\).

It follows that if an antiderivative \(F\) of a function \(f\) is known over an interval \(I\), then, any other antiderivative of \(f\) over \(I\) is of the form \(F(x)+C\).

It follows that for each differentiation rule, we have a rule to find the most general antiderivative of the derivative over an open interval.

Example 2.
(a) Since \(\displaystyle \frac{d}{dx} x^n = nx^{n-1}\), then, an antiderivative of \(x^n\), \(n\neq -1\), is \(\displaystyle \frac{1}{n+1}x^{x+1}+C\).

(b) Since \(\displaystyle \frac{d}{dx} \ln |x| = \frac{1}{x}\), then, an antiderivative of \(\displaystyle \frac{1}{x}\) is \(\displaystyle\ln |x|+C\).

Proposition 1.
Let \(F\) and \(G\) be antiderivatives of \(f\) and \(g\), respectively, and let \(k\) be any real number. Then

1. \(F+G\) is an antiderivative of \(f+ g\).
2. \(kF\) is an antiderivative of \(kf\).

Proof.
This is immediate from the analogous properties for derivatives:

1. \((F+G)’ = F’+G’ = f+ g\).

2. \((kF)’ = kF’ = kf\).

Example 3. Find an antiderivative of of the function (a) \(\displaystyle h(t) = \frac{t^3+1}{(t+1)\sqrt[3]{t^2}}\);    (b) \(f(x) = \tan^2 \pi x\).

Solution. (a) First, we simplify the function:
\[h(t) = \frac{t^3+1}{(t+1)\sqrt[3]{t^2}}= \frac{(t+1)(t^2-t+1)}{(t+1) t^{2/3}} = t^{4/3}-t^{1/3} + t^{-2/3},\]
provided \(t\neq -1\). Thus
\[H(t) = \frac{1}{7/3}t^{7/3} – \frac{1}{4/3}t^{4/3}+ \frac{1}{1/3}t^{1/3}+ C
= \frac{3}{7}t^2\,\sqrt[3]{t} – \frac{3}{4}t\,\sqrt[3]{t }+3\sqrt[3]{t }+ C
=\!\frac{3}{28}\sqrt[3]{t} \left (12t^2-21t+84\right ) +C.\]

(b) Since \(\tan^2 \pi x = \sec^2 \pi x -1\) and \(\displaystyle \frac{1}{\pi}\tan \pi x \) is an antiderivative of \( \sec^2 \pi x\), it follows that an antiderivative of \(f\) is
\(\displaystyle F(x) = \frac{1}{\pi}\tan \pi x -x +C \).

The above example shows that to identify the antiderivative of a function, it may be necessary to rewrite the function. Rewrite and simplify are two basic skills needed to find antiderivatives.

Example 4. Find the most general antiderivative of the function (a) \( \displaystyle f(x) =\frac{3}{\sin^2 \pi x}-2\frac{\csc^23x}{\sec 3x}\);    (b) the hyperbolic cosine, \( \displaystyle \cosh(t) = \frac{e^t+e^{-t}}{2}\).

Solution.
(a) First, we rewrite the function:
\[f(x) =\frac{2}{\sin^2 \pi x}+ 2\frac{\csc^23x}{\sec 3x} = 3 \csc^2 \pi x-2\csc 3x \frac{\csc 3x}{\sec 3x} = 3 \csc^2 \pi x-2\csc 3x \cot 3x.\]
Using Proposition 1, \(\displaystyle \frac{d}{dx}\cot a x = -a \csc^2ax\) and \(\displaystyle \frac{d}{dx}\csc a x = -a \csc ax \cot ax\), \(a \in \mathbb{R}\),
\[F(x) = – \frac{3}{\pi}\cot \pi x + \frac{2}{3} \csc 3x+C.\]
(b) Since an antiderivative of \(e{ax}\), \(a\neq 0\), is \(\displaystyle \frac{1}{a}e^{ax}+C\), an antiderivarive of hyperbolic cosine is
\[\frac{e^t-e^{-t}}{2} = \sinh t,\]
that is, hyperbolic sine.

Example 5. Find the most general antiderivative of the function \( \displaystyle g(t) = \frac{4+t^2}{1+t^2} – \frac{2}{\sqrt{1-t^2}}\).

Solution.
First, we rewrite the function:
\[g(t) = \frac{4+t^2}{1+t^2} – \frac{2}{\sqrt{1-t^2} }= \frac{3+ 1+t^2}{1+t^2} – 2\frac{1}{\sqrt{1-t^2}}
= 3\frac{1}{1+t^2} + 1 – 2\frac{1}{\sqrt{1-t^2}} \]
\[\Longrightarrow G(t) = 3\arctan t + t -2\arcsin t+C.\]

Example 6. Find antiderivatives of the squares of the six trigonometric functions.

The double angle identities play a key role when evaluating the antiderivatives of the squares of sine and cosine.

\[\cos 2x= \cos^2x-\sin^2x = 2\cos^2x-1= 1-2\sin^2x\hspace{.5in}\sin 2x = 2\sin x \cos x,\]
and
\[
\sin^2x = \frac{1-\cos 2x}{2}
\hspace{.5in}
\cos^2x = \frac{1+\cos 2x}{2}
\]

Solution.

\(\displaystyle h(x) = \sin^2x = \frac{1}{2} – \frac{\cos 2x}{2} \Longrightarrow H(x) = \frac{1}{2}x – \frac{\sin 2x}{4} +C\).

\(\displaystyle h(x) = \cos^2x = \frac{1}{2} + \frac{\cos 2x}{2} \Longrightarrow H(x) = \frac{1}{2}x + \frac{\sin 2x}{4} +C\).

\(h(x) = \sec^2x \Longrightarrow H(x) = \tan x +C\).

\(h(x) = \tan^2x = \sec^2x-1 \Longrightarrow H(x) = \tan x -x+C\).

\(h(x) = \csc^2x \Longrightarrow H(x) = -\cot x +C\).

\(h(x) = \cot^2x = \csc^2x-1 \Longrightarrow H(x) = -\cot x -x+C\).

Example 7. Find the most general antiderivative of the function.

1. \(\displaystyle h(x) =\frac{\cos 8x}{\cos 4x + \sin 4x}\);
2. \(\displaystyle h(x) =\frac{\sin 6x}{\cos^3 3x}\);
3. \(\displaystyle h(x) = \cos^2 \pi x\);
4. \(\displaystyle h(x) = \cos^4 \pi x – \sin^4\pi x\).

Solution. (a) Since
\[ \frac{\cos 8x}{\cos 4x + \sin 4x}
= \frac{\cos (2\cdot 4x)}{\cos 4x + \sin 4x}
= \frac{\cos^24x – \sin^2 4x}{\cos 4x + \sin 4x} = (\cos 4x – \sin 4x),\]

then, the most general antiderivative is
\[H(x) = \frac{1}{4}\sin 4x + \frac{1}{4}\cos 4x + C.\]

(b) Since
\[ h(x) = \frac{\sin 6x}{\cos^3 3x}
= \frac{\sin (2\cdot 3x)}{\cos^3 4x }
= \frac{2\sin 3x \cos 3x}{\cos^3 3x} = \frac{2\sin 3x }{\cos^2 3x} = 2\sec 3x \tan 3x\]
then, the most general antiderivative is
\[H(x) = \frac{2}{3}\sec 3x + C.\]

(c)
\[h(x) = \cos^2 \pi x = \frac{\cos 2 \pi x +1}{2}
\underset{?}{\Longrightarrow} H(x) = \frac{1}{2} \left (\frac{1}{2}\sin 2x + x\right )+C\]

(d)
\begin{align*}
h(x) &= \cos^4 \pi x – \sin^4\pi x = \left (\cos^2 \pi x – \sin^2\pi x\right ) \left( \cos^2 \pi x + \sin^2\pi x\right )= \cos^2 \pi x – \sin^2\pi x = \\
&=\cos 2\pi x \;\; \Longrightarrow \;\; H(x) =\frac{1}{2\pi}\sin 2\pi x+ C.
\end{align*}

Example 8. Determine whether the statement is true. If the statement is not true, provide an example showing that it fails.

(a) If \(F\) and \(G\) are antiderivatives of \(f\) and \(g\) respectively, then \(F\cdot G\) is an antiderivative of \(f\cdot g\).

(b) If \(F\) is an antiderivative of \(f\), then \(F(ax)\) is an antiderivative of \(f(ax)\), where \(a\) is a real number.

Solution.
(a) False. Since, by the Product Rule
\[(F\cdot G)’= F’\cdot G+F \cdot G’ = fG+Fg\;\; \Longrightarrow F G \;\; \text{ is an antiderivative of } \;\; fG+Fg.\]
For an example where the statement is false, let \(F(x) =x^2\) and \(G(x)=x^3\), and let \(f(x)=2x\) and \(g(x) =3x^2\). Then \((FG)(x) = x^5\), while \((fg)(x)= 6x^3\). Clearly, \((F\cdot G)’\neq fg\).

(b) False. By the Chain Rule
\[\frac{d}{dx}F(ax) = aF'(ax) = af(ax) \;\; \Longrightarrow \;\; F(ax) \;\;\text{ is an antiderivative of }\;\;af(ax).\]
For an example where the statement is false, let \(\displaystyle f(x) =\frac{1}{1+x^2}\), \(F(x)=\arctan x\), and \(a=2\). Then
\[F(2x) =\arctan 2x \;\;\underset{?}{\Longrightarrow}\;\; F'(2x) = \frac{2}{1+4x^2}\neq f(2x)=\frac{1}{1+4x^2}.\]

5.1.5 Differential Equations

A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation
\[f'(x) = g(x)\]
is a simple example of a differential equation. Solving this equation means finding a function \(f\) whose derivative is \(g\) . Therefore, the solutions of this equation are the antiderivatives of \(g\) . If \(G\) is one antiderivative of \(g\) , every function of the form \( G(x) + C\) is a solution of that differential equation.

Sometimes, we are interested in finding a solution to a differential equation \(y'(x)=f(x)\) that satisfies an additional condition of the form \(y(x_0) = y_0\). This type of problem is called an initial-value problem. The condition \(y(x_0) = y_0\) is known as an initial condition. Initial condition problems are particularly significant in applications of differential equations.

Example 1.
Solve the initial value problem \(y’=6t^2-2\), \(y(1) = -1\).

Solution.
First, solving the equation \(y’=6t^2-2\) yields \(y(t) = 2t^3-2t +C\). Next, we need to determine the value of \(C\) for which the initial condition \(y(1) = -1\) is satisfied:
\[y(1) =-1 \;\; \Longleftrightarrow \;\; 2-2+C=-1\;\; \Longleftrightarrow \;\; C=-1 \;\; \Longrightarrow y= 2t^3-2t -1.\]

Example 2.
Solve the initial value problem \(y”=\sin \pi x\), \(\displaystyle y\left (\frac{1}{2}\right ) =1\), \(\displaystyle y’\left (\frac{1}{2}\right ) = -2\).

Solution.
Thinking of \(y”=\sin \pi x\) as \((y’)’=\sin \pi x\), first we solve for \(y’\) and then for \(y\).

\[(y’)’=\sin \pi x \;\; \Longrightarrow \;\; y’ = -\frac{1}{\pi}\cos \pi x + C_1.\]
Then
\[y’\left (\frac{1}{2}\right ) = -2 \;\; \Longrightarrow \;\; -\frac{1}{\pi}\cos \left ( \pi \cdot\frac{1}{2} \right )+ C_1= -2\;\; \Longrightarrow C_1=-2 \;\;\Longrightarrow y’ = -\frac{1}{\pi}\cos \pi x -2.\]
We now solve for \(y\).
\[y = -\frac{1}{\pi^2}\sin \pi x -2x+C_0.\]
Then, solving for the remaining initial condition
\[y\left (\frac{1}{2}\right ) =1 \;\;\Longrightarrow \;\; -\frac{1}{\pi^2}\sin \left (\pi \cdot\frac{1}{2}\right ) -2\cdot\frac{1}{2} + C_0 = 1\;\; \Longrightarrow\;\; C_0 = 2 +\frac{1}{\pi^2}\;\; \Longrightarrow\]
\[\Longrightarrow\;\;
y = -\frac{1}{\pi^2}\sin \pi x -2x+2 +\frac{1}{\pi^2}.\]

Example 3.
Solve the boundary value problem \(f”(x)=8 e^{2x}\), \(f(1)=1\), \(f(2)=0\).

Solution.
\[f'(x) =4e^{2x} + c_1\Longrightarrow f(x) = 2 e^{2x} +c_1x + c_2.\]
Then
\[\hspace{.8in}\hphantom{(*)}f(1) = 2e^{2} + c_1 + c_2 = 1 \Longleftrightarrow
c_1 + c_2 = 1-2 e^{2},\hspace{.8in}(*)\]
and
\[\hspace{.8in}\hphantom{(**)}
f(2) = 2e^{4} + 2c_1 + c_2 = 0 \Longleftrightarrow
2c_1 + c_2 = – 2e^{4}. \hspace{.8in}(**)\]
Subtracting \((*)\) from \((**)\),
\[c_1=-1+2e^2-2e^4, \underset{(*)}{\Longrightarrow} c_2= 1-2e^2- c_1 = 1-2e^2-(-1+2e^2-2e^4) = 2 -4e^2+2e^4.\]
Thus
\[f(x) =2 e^{2 x}+(-1+2e^2-2e^4)x + ( 2 -4e^2+2e^4).\]

5.1.7 Applications to Rectilinear Motion

Let \(x=x(t)\) denote the position of a particle, at time \(t\), traveling along a straight line. Then, its velocity and acceleration are given, respectively by
\[v=v(t) = \frac{d}{dt}x, \;\;\; a = a(t) = \frac{d^2}{dt^2}x = \frac{d}{dt}v.\]
It follows that in this context, having information about the velocity or the acceleration plus initial conditions will lead to information about the position function.

Example 1.
Consider a particle that travels along a straight line with acceleration \(a= 2t+1\) m/s\(^2\). Assume further that at time \(t=1\) the particle is located at \(x(1)= 3\) m, with velocity \(v(1) =4\) m/s. Find the velocity and position of the particle at any time \(t\).

Solution. Since \(v’=a\), we have
\[v’=2t+1 \Longrightarrow v = t^2+t +v_0\Longrightarrow v(1) = 4 \Longleftrightarrow 1+1 + v_0= 4 \Longrightarrow v= t^2+t +2.\]
Since \(x’=v\), then
\[x’= t^2+t +2 \Longrightarrow x = \frac{1}{3}t^3+\frac{1}{2} t^2+2t +x_0 \Longrightarrow x(1) = \frac{1}{3}+\frac{1}{2}+2 +x_0= 3\Longleftrightarrow x_0 = \frac{1}{6} \Longrightarrow \]
\[\Longrightarrow x=\frac{1}{3}t^3+\frac{1}{2} t^2+2t +\frac{1}{6}. \]

5.1.9 Antiderivatives and the Chain Rule

As we have seen, each differentiation rule can be reinterpreted as an antiderivative rule. For example, let \(F(x) = \arcsin x + C\) and \(\displaystyle f(x) = \frac{1}{\sqrt{1-x^2}}\), then
\[F'(x) = f(x) \;\; \Longrightarrow \;\; F(x) \; \text{is the most general antiderivative of }\; f(x).\]

Here we consider examples related with the chain rule.
\[(f\circ g)'(x) = f'(g(x))g'(x) \]
It follows that
\[ (f\circ g)(x) +C \; \text{is the most general antiderivative of } \; f'(g(x))g'(x).\]
Thus, in order to use the chain rule in the context of antiderivatives, it is important to identify \(f'(x)\), \(g(x)\), \(g'(x)\), and \(f(x)\), to produce the antiderivative \(f(g(x))+C\).

Example 1.
Find the most general antiderivative of the function. (a) \(h(x) = 2x\cos \left ( x^2\right )\);   (b) \(\displaystyle h(x) = \frac{x+ \arccos^2 (2x)}{\sqrt{1 – 4x^2}}\);   (c) \(\displaystyle h(x) = e^{\cot 2x}{\csc^22x}\);   (d) \(\displaystyle h(x) = \frac{\ln^2 x}{x}\);   (e) \(\displaystyle h(x) = e^{\cos^2 x}\sin2 x\).

Solution. (a) Note that \(g(x)=x^2\), \(g'(x) = 2x\). Hence \(f'(x) = \cos x\). Thus \(f(x) = \sin x +C\) and the most general antiderivative of \(2x\cos \left ( x^2\right )\) is \(f(g(x))+C = \sin(x^2) +C\).

(b) In this example, it is necessary to do two things:

1. split the function into two separate functions;
2. adjust by a constant factor.

We split \(h\) as follows: \(\displaystyle h(x) = \frac{x+ \arccos^2 (2x)}{\sqrt{1 – 4x^2}} =
\frac{x }{\sqrt{1 – 4x^2}} + \frac{ \arccos^2 (2x)}{\sqrt{1 – 4x^2}}\).

Set \(\displaystyle h_1(x) = \frac{x }{\sqrt{1 – 4x^2}} \), and let \(g(x) = 1- 4x^2\), then \(g'(x) = -8x\). Thus, we adjust by a constant factor \(h_1\) as follows:
\[
\frac{x }{\sqrt{1 – 4x^2}} = \color{red}{ \pmb{ -\frac{1}{8}}} \frac{\pmb{ \color{red}{-8}}x}{\sqrt{1-4x^2}}. \Longrightarrow g(x) = 1-4x^2, \;\; g'(x) = -8x,\;\; f'(x) = -\frac{1}{8} x^{-1/2} \Longrightarrow \]
\[\Longrightarrow f(x) = -\frac{1}{4} x^{1/2}+C \Longrightarrow H_1(x) = -\frac{1}{4} (g(x))^{1/2}+C = -\frac{1}{4} (1-4x^2)^{1/2}+C = -\frac{1}{4} \sqrt{1-4x^2}+C. \]

Hence, \(\displaystyle H_1(x) = -\frac{1}{4} \sqrt{1-4x^2}+C\) is the most general antiderivative of \(\displaystyle h_1(x) = – \frac{x}{\sqrt{1-4x^2}}\).

Set \(\displaystyle h_2(x) = \frac{\arccos^22x }{\sqrt{1 – 4x^2}} \), and let \(g(x) =\arccos 2x\), then \(g'(x) =-\frac{2}{\sqrt{1 – 4x^2}} \). Thus, we adjust by a constant factor \(h_2\) as follows:
\[
\frac{\arccos^22x }{\sqrt{1 – 4x^2}} = \color{red}{ \pmb{-\frac{1}{2}}} \cdot \frac{ \color{red}{ \pmb{-2}}\arccos^22x }{\sqrt{1 – 4x^2}}. \Longrightarrow g(x) = \arccos 2x, \;\, g'(x) =-\frac{2}{\sqrt{1 – 4x^2}},\;\, f'(x) = -\frac{1}{2} x^2 \Longrightarrow \]
\[\Longrightarrow f(x) =- \frac{1}{6} x^3+C \Longrightarrow H_2(x) = -\frac{1}{6} (g(x))^3+C = -\frac{1}{6} (\arccos 2x)^3+C = -\frac{1}{6} \arccos^32x+C. \]

Hence, \(\displaystyle H_2(x) = -\frac{1}{6} \arccos^32x+C\) is the most general antiderivative of \(\displaystyle h_2(x) = \frac{\arccos^22x}{\sqrt{1-4x^2}}\).

The most general antiderivative of \(\displaystyle h(x) = \frac{x+ \arccos^2 (2x)}{\sqrt{1 – 4x^2}}\) is
\[ H(x) = -\frac{1}{4} \sqrt{1-4x^2} -\frac{1}{6} \arccos^32x+C.\]

(c) Let \(g(x) = \cot 2x\), then \(g'(x) = -2\csc^22x\). First, we adjust by a constant factor:

\[h(x) = e^{\cot 2x}{\csc^22x} = \pmb{-\frac{1}{2}} e^{\cot 2x}{\pmb{(-2)}\csc^22x}. \Longrightarrow g(x) = \cot 2x, \;\; g'(x) = -2\csc^2 2x, \;\; \Longrightarrow \]
\[\Longrightarrow f'(x) =-\frac{1}{2} e^x \Longrightarrow f(x) =-\frac{1}{2} e^x+C \Longrightarrow H(x) =-\frac{1}{2}e^{\cot 2x } + C.\]

(d) Let \(g(x) = \ln x\), then \(\displaystyle g'(x) = \frac{1}{x}\). And, \(f'(x) = x^2\). Thus
\[f(x) = \frac{1}{3} x^3 +C. \Longrightarrow H(x) = \frac{1}{3} \ln^3 x +C.\]

(e) Let \(g(x)= \cos^2x\) then \(g'(x) = -2\cos x \sin x \). Hence
\[h(x) = e^{\cos^2 x}\sin2 x = -e^{\cos^2 x}(-2\sin x\cos x). \Longrightarrow\]
\[\Longrightarrow f'(x) = -e^x \Longrightarrow f(x) = – e^x+C.\]
Thus, the most general antiderivative of \(\displaystyle h(x) = e^{\cos^2 x}\sin2 x\) is \(\displaystyle H(x) = – e^{\cos^2 x}+C\).