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Chapter 6: Problem 36

Determine each indefinite integral. \(\int \sinh ^{2} x d x\) (Hint: Use an identity.)

Short Answer

Expert verified
Question: Find the indefinite integral of \(\sinh^2 x\), given the hint that an identity can be used to simplify the integrand before integration. Answer: \(\int \sinh^2 x \, dx = \frac{1}{2}(\cosh x\sinh x) - x + C\)

Step by step solution

01

Recall the identity of hyperbolic functions

Recall that one of the identities for hyperbolic functions is: \(\cosh^2 x - \sinh^2 x = 1\) We can rearrange the formula to isolate \(\sinh^2 x\): \(\sinh^2 x = \cosh^2 x - 1\)
02

Substitute the integrand with the identity

Replace the integrand with the identity derived in Step 1: \(\int \sinh^2 x \, dx = \int (\cosh^2 x - 1) \, dx\)
03

Integrate each term separately

We can break up the integral and integrate each term separately: \(\int (\cosh^2 x - 1) \, dx = \int \cosh^2 x \, dx - \int 1 \, dx\)
04

Integrate the first term

To integrate the first term, \(\int \cosh^2 x \, dx\), recall that the derivative of \(\sinh x\) is \(\cosh x\). Thus, we can use integration by parts, with \(u = \cosh x\) and \(dv = \cosh x \, dx\): $$ \int \cosh^2 x \, dx = u \int dv - \int \big(\frac{du}{dx} \int dv\big)dx = \cosh x\sinh x - \int (\sinh x)(\sinh x \, dx) $$ Notice that \(\int (\sinh x)(\sinh x \, dx) = \int \sinh^2 x \, dx\), which we can call \(I\): \(I = \cosh x\sinh x - I\) Now, solve for \(I\): \(2I = \cosh x\sinh x \Rightarrow I = \frac{1}{2}(\cosh x\sinh x)\)
05

Integrate the second term

The integration of the second term is straightforward: \(\int 1 \, dx = x\)
06

Combine the results

Now, combine the results from Steps 4 and 5: \(\int \sinh^2 x \, dx = \int (\cosh^2 x - 1) \, dx = \int \cosh^2 x \, dx - \int 1 \, dx = \frac{1}{2}(\cosh x\sinh x) - x + C\) Where \(C\) is the constant of integration. So the final answer is: \(\int \sinh^2 x \, dx = \frac{1}{2}(\cosh x\sinh x) - x + C\)

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions, similar to trigonometric functions, are relationships that arise in hyperbolic geometry. They include functions such as sinh (hyperbolic sine) and cosh (hyperbolic cosine). These functions are defined using exponential functions:
  • The hyperbolic sine function, \( \sinh x \), is defined as \( \sinh x = \frac{e^x - e^{-x}}{2} \).
  • The hyperbolic cosine function, \( \cosh x \), is given by \( \cosh x = \frac{e^x + e^{-x}}{2} \).
One important identity involving hyperbolic functions is \( \cosh^2 x - \sinh^2 x = 1 \). This identity is similar to the Pythagorean identity used in trigonometry. It can be rearranged to express \( \sinh^2 x \) in terms of \( \cosh^2 x \), which becomes \( \sinh^2 x = \cosh^2 x - 1 \). This rearrangement is very useful in solving integrals involving hyperbolic functions.
Integration by Parts
Integration by parts is a useful technique for finding integrals, based on the product rule for differentiation. When faced with a product of functions, integration by parts can help break down the integral into more manageable parts. The formula is given as:\[ \int u \, dv = uv - \int v \, du \]
  • Choose \( u \) and \( dv \) wisely. Let \( u \) be a function that becomes simpler when differentiated and \( dv \) a function whose integral is easily determined.
  • The derivative \( du \) is found from \( u \), and \( v \) is the integral of \( dv \).
In this particular exercise, to integrate \( \cosh^2 x \), we marked \( u = \cosh x \) and \( dv = \cosh x \, dx \). Differentiating and integrating these respectively gives us the nice simplification to compute the integral effectively.
Integral Identities
Integral identities are essential in simplifying expressions within integrals. They provide a way to transform a complex integrand into a form that is easier to integrate. The hyperbolic identity \( \sinh^2 x = \cosh^2 x - 1 \) is an example of using an identity to simplify \( \int \sinh^2 x \, dx \). By substituting \( \sinh^2 x \) with \( \cosh^2 x - 1 \), the problem turns from one that seems complex into simpler integrals.
  • Identity substitutions can greatly reduce the difficulty by transforming the original integral.
  • Using identities can split the integral into parts that are more directly integrable.
This approach of applying identities allows mathematicians and students to solve problems that would otherwise require more complex methods.

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Most popular questions from this chapter

Verify the following identities. $$\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y$$

a. The definition of the inverse hyperbolic cosine is \(y=\cosh ^{-1} x \Leftrightarrow x=\cosh y,\) for \(x \geq 1,0 \leq y<\infty.\) Use implicit differentiation to show that \(\frac{d}{d x}\left(\cosh ^{-1} x\right)=\) \(1 / \sqrt{x^{2}-1}.\) b. Differentiate \(\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})\) to show that \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=1 / \sqrt{x^{2}+1}.\)

Consider the functions \(f(x)=a \sin 2 x\) and \(g(x)=(\sin x) / a,\) where \(a>0\) is a real number. a. Graph the two functions on the interval \([0, \pi / 2],\) for \(a=\frac{1}{2}, 1\) and 2. b. Show that the curves have an intersection point \(x^{*}\) (other than \(x=0)\) on \([0, \pi / 2]\) that satisfies \(\cos x^{*}=1 /\left(2 a^{2}\right),\) provided \(a>1 / \sqrt{2}\) c. Find the area of the region between the two curves on \(\left[0, x^{*}\right]\) when \(a=1\) d. Show that as \(a \rightarrow 1 / \sqrt{2}^{+}\). the area of the region between the two curves on \(\left[0, x^{*}\right]\) approaches zero.

Consider the cubic polynomial \(f(x)=x(x-a)(x-b),\) where \(0 \leq a \leq b\) a. For a fixed value of \(b,\) find the function \(F(a)=\int_{0}^{b} f(x) d x\) For what value of \(a\) (which depends on \(b\) ) is \(F(a)=0 ?\) b. For a fixed value of \(b\), find the function \(A(a)\) that gives the area of the region bounded by the graph of \(f\) and the \(x\) -axis between \(x=0\) and \(x=b\). Graph this function and show that it has a minimum at \(a=b / 2\). What is the maximum value of \(A(a),\) and where does it occur (in terms of \(b\) )?

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=e^{-x}, y=0, x=0,\) and \(x=p>0\) revolved about the \(x\) -axis (Is the volume bounded as \(p \rightarrow \infty ?\))
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