Question

Evaluate the following integrals. Include absolute values only when needed. $$\int \frac{4^{\cot x}}{\sin ^{2} x} d x$$

Short answer

Question: Evaluate the following integral: $$\int \frac{4^{\cot x}}{\sin ^{2} x} dx $$ Answer: The integral can be evaluated as $$\int \frac{4^{\cot x}}{\sin ^{2} x} dx = -\frac{2}{\ln 4}e^{\cot x \ln 4} + C$$

Step-by-step solution

Simplify the integrand

Observe that the given integral can be simplified using the trigonometric identity \(\sin^{2}(x) + \cos^{2}(x) = 1\). If we rewrite the given integral using this identity, we have: $$\int \frac{4^{\cot x}}{\sin ^{2} x} dx = \int \frac{4^{\frac{\cos x}{\sin x}}}{1-\cos^2 x} dx $$

Perform a substitution

Let's perform a change of variable: \(u=\cos x\), which implies \(du=-\sin xdx\). Now, rearrange the trigonometric identity to get \(\sin^{2}(x) = 1-\cos^{2}(x) = 1-u^{2}\). Also, since \(-\sin xdx = du\), we can replace \(\sin xdx = -du\). Therefore, the integral transforms as: $$\int \frac{4^{\cot x}}{\sin ^{2} x} dx = \int \frac{4^{\frac{\cos x}{\sin x}}}{1-\cos^2 x} dx = -\int \frac{4^{\frac{u}{\sqrt{1-u^2}}}}{1-u^2} du $$

Integrate

Now, we can integrate the new function, $$-\int \frac{4^{\frac{u}{\sqrt{1-u^2}}}}{1-u^2} du $$ Let \(v=\frac{u}{\sqrt{1-u^2}}\). Then, differentiating both sides with respect to \(u\), we find that \(dv = \frac{du}{(1-u^2)^{\frac{3}{2}}}\). Replacing \(4^{\cot x}\) with \(e^{\ln 4^{\frac{u}{\sqrt{1-u^2}}}} = e^{\frac{u}{\sqrt{1-u^2}}\ln 4}\) and rewriting the integral, we get: $$-\int \frac{e^{v\ln 4}}{(1-u^2)\frac{dv}{du}} du$$ Now, substitute \(v\) back with \(\frac{u}{\sqrt{1-u^2}}\), and \(dv\) with \(\frac{du}{(1-u^2)^{\frac{3}{2}}}\): $$-\int \frac{e^{\frac{u}{\sqrt{1-u^2}}\ln 4}}{(1-u^2)\frac{du}{(1-u^2)^{\frac{3}{2}}}} du = -\int e^{\frac{u}{\sqrt{1-u^2}}\ln 4}(1-u^2)^{\frac{1}{2}} du$$ Using the substitution method, we now integrate the function, which results in: $$-\int e^{\frac{u}{\sqrt{1-u^2}}\ln 4}(1-u^2)^{\frac{1}{2}} du = -\frac{2}{\ln 4}e^{\frac{u}{\sqrt{1-u^2}}\ln 4} + C$$

Resubstitute the original variable

Now we will reverse the substitution we performed earlier, first replacing \(u\) with \(\cos x\) and then substituting \(\cot x\): $$ -\frac{2}{\ln 4}e^{\frac{\cos x}{\sqrt{1-\cos^2 x}}\ln 4} + C = -\frac{2}{\ln 4}e^{\cot x \ln 4} + C$$ This is the final answer for the given integral: $$\int \frac{4^{\cot x}}{\sin ^{2} x} dx = -\frac{2}{\ln 4}e^{\cot x \ln 4} + C$$

Key concepts

Trigonometric identities

Understanding trigonometric identities is crucial in calculus and algebra. They provide us with relationships between different trigonometric functions, like sine and cosine. One of the most fundamental identities is \(\sin^{2}(x) + \cos^{2}(x) = 1.\)
This identity helps simplify expressions and integrate complex equations.

• In the original exercise, this identity was used to transform the denominator from \(\sin^{2}(x)\) into \(1-\cos^{2}(x)\).
• Rewriting expressions using trigonometric identities makes them easier to manipulate and solve.

By mastering these identities, you can simplify and solve many trigonometric integrals more effectively.

Substitution method

The substitution method is a powerful tool in integration. It involves changing the variable of integration to simplify an integral. In the given problem, the substitution \(u = \cos x\) was made.
This leads to \(du = -\sin x \, dx\), altering the entire integral.

• Substitution helps in transforming complex expressions into simpler forms that can be easily integrated.
• It's essential to also adjust the differential \(dx\) to match the substitution by calculating \(du\).

Practicing substitution with various functions increases your ability to solve complex integrals.

Integral transformation

Integral transformation involves altering the form of an integral to make it more manageable. This can include manipulating both the integrand and the limits (in the case of definite integrals).
The problem used transformations such as expressing \(4^{\cot x}\) as \(e^{\cot x \ln 4}\) to further simplify and prepare for integration.

• Transformations allow you to apply integration techniques, like exponential integration, more easily.
• Changing expressions into exponential forms can reveal simpler solutions.

Understanding how to shift between different integrand forms is critical in problem-solving.

Definite integrals

In the exercise at hand, we focused on indefinite integration, but understanding definite integrals is key for calculating actual numerical values of integrals over specific intervals.
Definite integrals have upper and lower limits, providing the exact area under a curve within those limits.

• By applying limits after integration, you transform the antiderivative into a specific solution.
• This concept is widely used in physics and engineering for calculating totals like areas, volumes, and other quantities.

Mastering definite integrals involves understanding both the antiderivatives and applying the fundamental theorem of calculus.