Question

Evaluate the following integrals. $$\int e^{x} \sin ^{998}\left(e^{x}\right) \cos ^{3}\left(e^{x}\right) d x$$

Short answer

In this problem, we used substitution and integration by parts methods to solve the given integral. First, we performed the substitution \(u = e^x\), which transformed the integral into terms of \(u\). Then, we applied integration by parts to evaluate the integral. Finally, we reverted back to the original variable \(x\) to get the solution, which is: $$\int e^{x} \sin^{998}(e^x) \cos^3(e^x) dx = \frac{\sin^{998}(e^x) \frac{1}{3} \cos^2(e^x) \sin(e^x)}{1+\frac{998}{3}} + C$$ Where \(C\) is the constant of integration.

Step-by-step solution

Substitution

Let's perform the substitution \(u = e^x\). Then, \(du = e^x dx\), and our integral becomes: $$\int e^{x} \sin^{998}(u) \cos^{3}(u) du$$ Now, the integral is in terms of \(u\).

Integration by Parts

In order to evaluate the integral, let's use integration by parts. Recall the formula for integration by parts: $$\int u dv = uv - \int v du$$ Let: $$u = \sin^{998}(u)$$ $$dv = \cos^3(u) du$$ Differentiate \(u\) to get \(du\), and antidifferentiate \(dv\) to get \(v\): $$du = 998 \sin^{997}(u) \cos(u) du$$ $$v = \frac{1}{3} \cos^2(u) \sin(u)$$ Then, using integration by parts, we can rewrite our integral as: $$\int u dv = uv - \int v du = \sin^{998}(u) \frac{1}{3} \cos^2(u) \sin(u) - \int \frac{1}{3} \cos^2(u) \sin(u)(998 \sin^{997}(u) \cos(u) du)$$

Simplify and Integrate

Now, let's simplify the remaining integral: $$\int \frac{1}{3} \cos^2(u) \sin(u)(998 \sin^{997}(u) \cos(u) du)= \frac{998}{3}\int \sin^{998}(u) \cos^3(u) du$$ Now we have the original integral on the right-hand side, so we can call this integral \(I\): $$I = \sin^{998}(u) \frac{1}{3} \cos^2(u) \sin(u) -\frac{998}{3}I$$ Now, we can solve for \(I\) as follows: $$I + \frac{998}{3}I =\sin^{998}(u) \frac{1}{3} \cos^2(u) \sin(u)$$ $$I= \frac{\sin^{998}(u) \frac{1}{3} \cos^2(u) \sin(u)}{1+\frac{998}{3}}$$

Revert the Substitution

Now that we have the integral in terms of \(u\), we need to revert back to \(x\). Recall that \(u = e^x\): $$I = \frac{\sin^{998}(e^x) \frac{1}{3} \cos^2(e^x) \sin(e^x)}{1+\frac{998}{3}}$$ Thus, the solution to the integral is: $$\int e^{x} \sin^{998}(e^x) \cos^3(e^x) dx = \frac{\sin^{998}(e^x) \frac{1}{3} \cos^2(e^x) \sin(e^x)}{1+\frac{998}{3}} + C$$ Where \(C\) is the constant of integration.

Key concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify integration. It involves changing variables to make integration easier. In the context of integration, we substitute a part of the integral with a new variable. This transforms the original integral into one that is simpler to solve.

In our given problem, the substitution is performed by setting \(u = e^x\). This makes \(du = e^x dx\).

• First, identify a part of the integral that can be substituted. In this case, \(e^x\) fits naturally, as \(u\) in place of \(e^{x}\) simplifies the expressions and matching the differential \(du\) allows us to transform the entire integral into the \(u\) variable.
• The new integral becomes \(\int e^{x} \sin^{998}(u) \cos^{3}(u) du\), where the presence of \(e^x\) cancels out nicely with \(du\).

This conversion is essential since it reduces complex expressions and helps in further manipulation using other integration techniques.

Trigonometric Functions

Trigonometric functions are fundamental in calculus, especially in integrations involving periodic oscillations or wave-like properties. In our exercise, we encounter \(\sin(u)\) and \(\cos(u)\), which are elevated to high powers, making direct integration challenging.

Understanding specific identities and manipulation techniques with trig functions is crucial. Here are some strategies that help:

• Simplifying products of sine and cosine using identities can sometimes transform the expression into a more workable form, allowing for clearer integration paths.
• Another practice is using powers of trigonometric functions, like \(\sin^a(u)\) and \(\cos^b(u)\), by breaking them into derivatives and simpler units if possible. This can be crucial when setting up integration by parts, as seen in our solution.

In this problem, you can see that the \(998^{th}\) power of \(\sin(u)\) and the third power of \(\cos(u)\) are dealt with using these advanced techniques, making it imperative to be familiar with handling high-powered trigonometric functions.

Exponential Functions

Exponential functions feature prominently in calculus due to their naturally occurring nature in growth and decay problems. In our integral, \(e^x\) plays an important role both as the substitution variable and within the transformed trigonometric functions.

Working with exponentials involves leveraging their properties:

• The function \(e^x\) is unique as its derivative and integral are the same, making it a versatile component in calculus.
• In substitutions, exponential functions offer a seamless transition between variables because of their differential \(e^x dx\), allowing other complex expressions to be simplified effectively.

In this context, \(e^x\) is substituted out to ease the integration process, but it comes back into play when reverting substitution. Understanding how exponentials transform within such calculus problems is key to mastering integration by parts involving these functions.