Question

Multiple substitutions If necessary, use two or more substitutions to find the following integrals. $$\begin{aligned} &\int x \sin ^{4} x^{2} \cos x^{2} d x\left(\text {Hint}: \text { Begin with } u=x^{2},\text { and then use }\right.\\\ &v=\sin u .) \end{aligned}$$

Short answer

Based on the step-by-step solution above, the integral of the function \(x \sin^4{x^2} \cos{x^2}\) with respect to \(x\) is given by: \[\int x \sin^4{x^2} \cos{x^2} dx = \frac{1}{10}\sin^5{x^2} + C\] where \(C\) represents the constant of integration.

Step-by-step solution

First Substitution (\(u = x^2\))

We are given the integral: $$\int x \sin^4{x^2} \cos{x^2} dx$$ We make the substitution \(u = x^2\). Then, the derivative of \(u\) with respect to \(x\) is: $$\frac{du}{dx} = 2x$$ Now we need to find \(dx\) in terms of \(du\). By dividing both sides by \(2x\), we get: $$dx = \frac{du}{2x}$$

Replace \(x\) and \(dx\) with \(u\)

Now, we replace the expression of \(x\) and \(dx\) with \(u\) using the substitutions and compute the new integral: $$\int x \sin^4{u} \cos{u} \frac{du}{2x}$$ The \(x\) terms cancel out, and we are left to integrate: $$\frac{1}{2}\int \sin^4{u} \cos{u} du$$

Second Substitution (\(v = \sin{u}\))

We make a second substitution, \(v = \sin{u}\). Then, the derivative of \(v\) with respect to \(u\) is: $$\frac{dv}{du} = \cos{u}$$ Now we need to find \(du\) in terms of \(dv\). By dividing both sides by \(\cos{u}\), we get: $$du = \frac{dv}{\cos{u}}$$

Replace \(u\) and \(du\) with \(v\)

Now, we replace the expression of \(u\) and \(du\) with \(v\) using the substitutions and compute the new integral: $$\frac{1}{2}\int v^4 \cos{u} \frac{dv}{\cos{u}}$$ The \(\cos{u}\) terms cancel out, and we are left to integrate: $$\frac{1}{2}\int v^4 dv$$

Integrate

Integrating \(v^4\) with respect to \(v\), we get: $$\frac{1}{2}\left[\frac{v^5}{5}\right] + C$$

Back-substituting

Now, we have to reverse the substitutions to get the final result. First, we back-substitute \(v\) with \(u\). So, $$\frac{1}{2}\left[\frac{(\sin{u})^5}{5}\right] + C$$ Then, we back-substitute \(u\) with \(x\). We get, $$\frac{1}{2}\left[\frac{(\sin{x^2})^5}{5}\right] + C$$ Therefore, the result of the integral is: $$\frac{1}{10}\sin^5{x^2} + C$$

Key concepts

Understanding the Substitution Method

The substitution method is an essential technique in integral calculus. It simplifies complex integrals by changing variables. This approach transforms the original integral into a simpler or more familiar form. Here's how the substitution method works:

• Identify a part of the integral to substitute (often something inside a function or with a known derivative).
• Replace this part with a new variable, say \( u \), and express \( du \) in terms of \( dx \).
• Transform the limits of integration if you are dealing with a definite integral.
• Solve the resulting integral in terms of \( u \).
• Finally, back-substitute the original variable to get your solution.

Using substitution reduces complexity by converting tricky expressions into more manageable forms, making calculus problems more approachable.

Integral Substitution: A Step-by-Step Approach

Integral substitution can sometimes involve multiple steps. In such cases, it’s crucial to tackle one substitution at a time. Consider the integral \( \int x \sin^4{x^2} \cos{x^2} \, dx \). Here's a breakdown:

First, substitute \( u = x^2 \). The derivative \( \frac{du}{dx} = 2x \) helps convert \( dx \) to \( \frac{du}{2x} \). After substitution, terms cancel and simplify to \( \frac{1}{2} \int \sin^4{u} \cos{u} \, du \).

Next, use the second substitution, \( v = \sin{u} \). Express \( du \) as \( \frac{dv}{\cos{u}} \), simplifying further to \( \frac{1}{2} \int v^4 \, dv \). Integrate and back-substitute to retrieve the original variable.

The process systematically converts the integral into a solvable format, highlighting the power of substitution in calculus.

Trigonometric Integrals Explained

Trigonometric integrals often involve powers of sine and cosine functions, like in \( \sin^4{x^2} \cos{x^2} \). Solving these integrals requires special strategies due to their periodic nature.

When dealing with trigonometric integrals:

• Look for substitution opportunities that exploit trigonometric identities (e.g., \( \sin^2x + \cos^2x = 1 \)).
• Consider power reduction or angle sum identities if applicable.
• Plan the substitution carefully to handle complex powers or products of trigonometric functions.

Understanding these principles allows you to break down and solve trigonometric integrals more efficiently. Simplifying using identities and substitutions makes seemingly intimidating problems manageable.