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Chapter 7: Problem 42

Find the integrals. Check your answers by differentiation. $$\int \frac{x \cos \left(x^{2}\right)}{\sqrt{\sin \left(x^{2}\right)}} d x$$

Short Answer

Expert verified
The integral is \( \sqrt{\sin(x^2)} + C \).

Step by step solution

01

Choose a Suitable Substitution

For the integral \( \int \frac{x \cos(x^2)}{\sqrt{\sin(x^2)}} \, dx \), notice that the expression inside the trigonometric functions is \( x^2 \). This suggests a substitution where \( u = x^2 \). Differentiating both sides gives \( du = 2x \, dx \), hence \( x \, dx = \frac{1}{2} \, du \). Substitute these into the integral.
02

Substitute and Simplify the Integral

After substituting, the integral becomes \( \int \frac{x \cos(x^2)}{\sqrt{\sin(x^2)}} \, dx = \int \frac{\cos(u)}{\sqrt{\sin(u)}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int \frac{\cos(u)}{\sqrt{\sin(u)}} \, du \). This integral is easier to tackle.
03

Perform Integration

Continue with the substitution method. Let \( v = \sin(u) \), which means \( dv = \cos(u) \, du \). The integral \( \frac{1}{2} \int \frac{\cos(u)}{\sqrt{\sin(u)}} \, du \) can be rewritten using this substitution as \( \frac{1}{2} \int \frac{1}{\sqrt{v}} \, dv \).
04

Solve the New Integral

The integral \( \frac{1}{2} \int \frac{1}{\sqrt{v}} \, dv \) is a standard integral form. The result is \( \frac{1}{2} \cdot 2 \sqrt{v} + C = \sqrt{v} + C \).
05

Back-Substitute to the Original Variable

Recall that \( v = \sin(u) \) and \( u = x^2 \). Substituting back, you get the result \( \sqrt{\sin(x^2)} + C \) where \( C \) is the integration constant.
06

Verify by Differentiation

Differentiate the result \( \sqrt{\sin(x^2)} + C \) to verify. Using the chain rule, \( \frac{d}{dx}(\sqrt{\sin(x^2)}) = \frac{1}{2\sqrt{\sin(x^2)}} \cdot \cos(x^2) \cdot 2x = \frac{x \cos(x^2)}{\sqrt{\sin(x^2)}} \), which matches the original integrand, verifying that the integral is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
When we encounter complex integrals, the substitution method becomes a handy tool. Essentially, it involves changing variables to simplify integration processes. For the given integral \( \int \frac{x \cos(x^2)}{\sqrt{\sin(x^2)}} \, dx \), the substitution method was utilized effectively.
Instead of dealing with the intricate form of \( x^2 \) under the cosine and sine functions, we engage in variable substitution:
  • Choose \( u = x^2 \). This choice simplifies the integral鈥檚 inner functions.
  • Differentiate the substituted variable: \( du = 2x \, dx \).
  • Solve for \( x \, dx \), which results in \( x \, dx = \frac{1}{2} \, du \).
Substituting these back transforms our integral into a simpler form, making it more manageable. This transition primarily aims to convert trigonometric functions into algebraic expressions, facilitating easier integration.
Definite Integral
In this solution, even though a definite integral wasn't exactly calculated, understanding this concept is crucial. A definite integral represents the net area under a curve over a defined interval. However, here we dealt with an indefinite integral, as no specific limits were provided.
When evaluating integrals in general, and particularly when using substitution, the process allows us to redefine the problem within new bounds if they were given:
  • After substituting, replace old bounds in terms of \( x \) with bounds in terms of \( u \).
  • Evaluate the integral in this new range.
  • Back-substitute to find the original variable's context.
By completing these steps with the inclusion of limits, it would offer the net result over a specific interval, yielding the area bounded by the curve's graph and the x-axis.
Trigonometric Functions
Trigonometric functions frequently appear in calculus problems, notably in integration exercises. In such contexts, understanding their interactions and transformations is vital. For the integral \( \int \frac{x \cos(x^2)}{\sqrt{\sin(x^2)}} \, dx \), trigonometric functions are deeply involved.
Let's break down their usage:
  • Both sine and cosine functions are expressed through the substitution \( x^2 \), highlighting the angle dependence typical in trigonometric contexts.
  • This integral, involving \( \cos(u) \) and \( \sin(u) \), showcases a typical use wherein trigonometric identities or simplifications facilitate integration.
  • Given their periodic nature, understanding amplitude, phase shift, and transformation valuable aids in anticipating integral results.
Embracing these trigonometric principles assists in learning how subtle changes in angles (like \( x^2 \)) impact function outcomes and integral solutions.

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Most popular questions from this chapter

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An electric current, \(I(t),\) flowing out of a capacitor, decays according to \(I(t)=I_{0} e^{-t},\) where \(t\) is time. Find the charge, \(Q(t),\) remaining in the capacitor at time \(t\) The initial charge is \(Q_{0}\) and \(Q(t)\) is related to \(I(t)\) by $$ Q^{\prime}(t)=-I(t) $$
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