Problem 32 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{4^{\cot x}}{\sin ^{2} x} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int \frac{4^{\cot x}}{\sin^2 x} dx\). Answer: \(\int \frac{4^{\cot x}}{\sin^2 x} dx = 4^{\cot x} \csc x + C\).

Step 1: Substitute for \(\cot x\) in terms of \(\sin x\) and \(\cos x\)

To find a more manageable expression, we will rewrite \(\cot x\) as \(\frac{\cos x}{\sin x}\). This will allow us to rewrite the integral as follows: $$ \int \frac{4^{\cot x}}{\sin^2 x} dx = \int \frac{4^{\frac{\cos x}{\sin x}}}{\sin^2 x} dx. $$

Step 2: Use the substitution \(u=\sin x\)

Now, we make the substitution \(u = \sin x\). This will simplify our integral and allow us to separate the \(dx\) term. Differentiating both sides with respect to \(x\), we get: $$ \frac{d u}{d x} = \cos x \Rightarrow d u = \cos x d x. $$ We can substitute this back into the integral: $$ \int \frac{4^{\frac{\cos x}{\sin x}}}{\sin^2 x} dx = \int \frac{4^{\frac{1}{u}}}{u^2} (\cos x dx) = \int \frac{4^{\frac{1}{u}}}{u^2} du, $$ where we used the fact that \(\cos x dx = du\).

Step 3: Apply integration by parts

To solve this integral, we can use integration by parts. Integration by parts is the process of decomposing an integral into the product of two functions, \(u\) and \(dv\), and solving for the integral using the following formula: $$ \int u dv = uv - \int v du. $$ Let \(u = 4^{\frac{1}{u}}\) and \(dv = -u^{-2} du\). Then \(du = -\frac{\ln 4}{u^2} 4^{\frac{1}{u}} du\) and \(v = u^{-1}\). Now, using the integration by parts formula: $$ \int \frac{4^{\frac{1}{u}}}{u^2} du = u^{-1} 4^{\frac{1}{u}} + \ln 4 \int u^{-1} 4^{\frac{1}{u}} du. $$

Step 4: Evaluate the remaining integral

Next, we will evaluate the remaining integral. Notice that the remaining integral is just a constant multiple of our original integral. Let \(I\) denote the original integral: $$ I = \int \frac{4^{\frac{1}{u}}}{u^2} du. $$ Then, the integral we must solve is: $$ \int u^{-1} 4^{\frac{1}{u}} du = \frac{1}{\ln 4} I. $$ Now, plug this back into the expression for \(I\): $$ I = u^{-1} 4^{\frac{1}{u}} + \ln 4(\frac{1}{\ln 4} I) \Rightarrow I = 4^{\frac{1}{u}} u^{-1} + I. $$ Now, we can subtract \(I\) from both sides: $$ I = 4^{\frac{1}{u}} u^{-1}. $$

Step 5: Reverse the substitution

Finally, since we made a substitution with \(u = \sin x\) in the beginning, we now reverse that substitution: $$ \int \frac{4^{\cot x}}{\sin^2 x} dx = 4^{\frac{1}{\sin x}} (\sin x)^{-1} + C = 4^{\cot x} \csc x + C, $$ where \(C\) is the constant of integration. So, the evaluated integral is: $$ \int \frac{4^{\cot x}}{\sin^2 x} dx = 4^{\cot x} \csc x + C.

Key Concepts

Substitution Method

The substitution method is a fundamental technique in calculus, especially useful in integration, for simplifying complex integrals. It involves choosing a substitution that transforms an integral into a simpler form. The key is to change the variable of integration to make the integral easier to evaluate.

For example, in the given problem, we substitute \(u = \sin x\). This substitution was chosen because it directly relates to simplifying the integral's structure by expressing it in terms of \(u\), which makes dealing with the exponential and trigonometric functions more manageable. Here are the steps to employing a substitution:

• Choose \(u\) such that it simplifies the expression. Often \(u\) is set equal to the inside function of a composition.
• Write \(du\) in terms of \(dx\) to change the variable of integration, linking \(du\) to the original differential.
• Substitute \(u\) and \(du\) into the original integral to transform it into a new integral solely in terms of \(u\).

This way, a complicated function of \(x\) can become a straightforward function of \(u\), making the integration process more feasible.

Integration by Parts

Integration by parts is another crucial technique in integration, which is especially helpful when dealing with the integral of the product of two functions. It is a powerful tool that relates the integral of a product to an integral of a derivative of that product. The principle stems from the product rule of differentiation.

The formula used for integration by parts is:\[\int u dv = uv - \int v du,\]where \(u\) and \(dv\) are parts of the original integral. In each integral, one part is differentiated to give \(du\), and the other is integrated to give \(v\).

In our solution, we selected \(u = 4^{1/u}\) and \(dv = -u^{-2} du\). This particular choice was made to reduce the integral to a manageable form by leveraging the exponential nature of \(4^{1/u}\).

• First, decide on \(u\) and \(dv\).
• Differentiate \(u\) to find \(du\).
• Integrate \(dv\) to find \(v\).
• Substitute these into the integration by parts formula.
• Evaluate the new integral, which should be simpler.

Integration by parts can sometimes be iterative or require further substitution, but it is invaluable for handling products and compositions that would otherwise be non-trivial to integrate.

Trigonometric Integrals

Trigonometric integrals involve integrating functions whose integrands contain trigonometric functions of the form \(\sin x\), \(\cos x\), \(\tan x\), and so on. These integrals often benefit from trigonometric identities, substitutions, and advanced techniques like integration by parts.

In the original exercise, the trigonometric function that plays a central role is \(\sin x\). The substitution \(u = \sin x\) was used to handle the complexity brought by the trigonometric components inside the integral. Here are some general strategies when dealing with trigonometric integrals:

• Use trigonometric identities to simplify the integrand when possible.
• Look for opportunities to use substitution to simplify integrals involving powers of trigonometric functions, as was done with \(\sin x\) in our integral.
• Apply symmetry properties of trigonometric functions to evaluate definite integrals.

By using these strategies, we can transform complicated trigonometric expressions into simpler ones, making them much more approachable. Trigonometric integrals require a good understanding of both basic calculus techniques and trigonometric identities to find solutions efficiently.