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Chapter 11: Problem 23

Evaluate the definite integral. $$ \int_{1}^{4} \sec ^{-1} \sqrt{x} d x $$

Short Answer

Expert verified
\( \frac{2\pi}{3} - 2 \ln(2 + \sqrt{3}) \)

Step by step solution

01

Simplify the integrand

Introduce a substitution to simplify the integration process. Let \( u = \sec^{-1}(\sqrt{x}) \). Hence, \( x = \sec^2(u) \) and \( dx = 2\sec(u) \tan(u) du \).
02

Change the limits of integration

Convert the limits of integration to the new variable \( u \). When \( x = 1 \), \( \sec^{-1}(\sqrt{1}) = 0 \). When \( x = 4 \), \( \sec^{-1}(\sqrt{4}) = \pi/3 \). Therefore, the new limits are from \( 0 \) to \( \pi/3 \).
03

Substitute and integrate

Substitute \( x \), \( dx \), and the limits into the integral:\[ \int_{0}^{\pi/3} u * 2\sec(u) \tan(u) du \]. This simplifies to \[ 2 \int_{0}^{\pi/3} u \sec(u) \tan(u) du \]. Use integration by parts, letting \( v = u \) and \( dw = 2 \sec(u) \tan(u) du \). Therefore, \( dv = du \) and \( w = 2\sec(u) \).
04

Apply integration by parts

Use the integration by parts formula \( \int u dv = uv - \int v du \). Applying this:\[ 2[ u\sec(u) ]_{0}^{\pi/3} - \int_{0}^{\pi/3} 2 \sec(u) du = 2[ u\sec(u) - \int \sec(u) du ]_{0}^{\pi/3} \].
05

Evaluate and simplify

Evaluate the remaining integrals:\[ 2[ u\sec(u) - \ln |\sec(u) + \tan(u)| ]_{0}^{\pi/3} \]. Plug in the bounds:\[ 2[ \frac{\pi}{3} \sec(\pi/3) - \ln |\sec(\pi/3) + \tan(\pi/3)| ] - 2[ 0 - \ln |\sec(0) + \tan(0)| ] \]. Simplifying gives \( 2[ \frac{\pi}{3}(2) - \ln |2 + \sqrt{3}| - 0 + \ln |1| ] = \frac{2\pi}{3} - 2 \ln(2 + \sqrt{3}) \).

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Key Concepts

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

Integration by parts
Integration by parts is a technique used to integrate the product of two functions. It is derived from the product rule for differentiation. The formula for integration by parts is: \(\int u \, dv = uv - \int v \, du\).
In the solution of the given problem, we use integration by parts to simplify the integral \[2 \int_{0}^{\pi/3} u \, \sec(u) \, \tan(u) \, du \].
We let \(u = u\) and \( dv = 2 \sec(u) \tan(u) \, du \). This choice leads to \(v = 2 \sec(u)\) and \(du = du\). Applying the integration by parts formula gives us
\[2 [ u \, \sec(u)]_{0}^{\pi/3} - \int_{0}^{\pi/3} 2 \sec(u) \, du \].
This helps by reducing the problem to simpler integrals that we can evaluate.
Substitution Method
The substitution method is a technique for simplifying an integrand by changing variables. This method often makes challenging integrals more manageable.
  • The first step is to choose an appropriate substitution.
  • Next, we convert all variables and differential elements to this new variable.
In our example, we start by letting \( u = \sec^{-1}(\sqrt{x}) \). As a result, \( x = \sec^{2}(u) \) and \( dx = 2 \sec(u) \tan(u) \, du \).
The integral becomes \[ \int_{1}^{4} \sec^{-1}(\sqrt{x}) \, dx = \int_{0}^{\pi/3} u \cdot 2 \sec(u) \tan(u) \, du \].
This simplification transforms the integral into a form where integration by parts can be applied.
Limits of Integration
When performing a substitution in a definite integral, it is crucial to adjust the limits of integration to match the new variable.
  • Convert the lower and upper bounds using the substitution relation.
In our example, the substitution \( u = \sec^{-1}(\sqrt{x}) \) changes the limits:
For \( x = 1 \rightarrow u = \sec^{-1}(\sqrt{1}) = 0 \).
For \( x = 4 \rightarrow u = \sec^{-1}(\sqrt{4}) = \frac{\pi}{3} \).
Therefore, the original integral \[ \int_{1}^{4} \sec^{-1}(\sqrt{x}) \, dx \] becomes \[ \int_{0}^{\pi/3} u \cdot 2 \sec(u) \tan(u) \, du \] with new limits from 0 to \(\pi/3\).
This step ensures that we integrate over the correct interval after substitution.

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Most popular questions from this chapter

Show that the exact value of \(\int_{0}^{2} \sqrt{4-x^{2}} d x\) is \(\pi\). Approximate the definite integral by the trapezoidal rule to three decimal places, with \(n=8\), and compare the value so obtained with the exact value.

Evaluate the indefinite integral. $$ \int \frac{\left(x^{2}+2 x-1\right) d x}{27 x^{3}-1} $$

A particle is moving along a straight line so that if \(v \mathrm{ft} / \mathrm{sec}\) is the velocity of the particle at \(t \mathrm{sec}\), then $$ v=\frac{t^{2}-t+1}{(t+2)^{2}\left(t^{2}+1\right)} $$ Find a formula for the distance traveled by the particle from the time when \(t=0\) to the time when \(t=t_{1}\).

Evaluate the indefinite integral. $$ \int \frac{d x}{16 x^{4}-8 x^{2}+1} $$

Evaluate the indefinite integral. $$ \int \frac{d x}{x \sqrt{x^{2}+4 x-4}} $$
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