Problem 45 - Calculus for Biology and Medicine

Question

In Problems \(39-48\), first make an appropriate substitution and then use integration by parts to evaluate each integral. $$ \int_{0}^{\pi / 4} \cos \sqrt{x} d x $$

Step-by-Step Solution

Answer

\( \frac{\sqrt{\pi}}{2} \sin\left(\frac{\sqrt{\pi}}{2}\right) + \cos\left(\frac{\sqrt{\pi}}{2}\right) - 1 \)

Step 1: Make a Substitution

Let's make the substitution \( u = \sqrt{x} \). This implies that \( x = u^2 \). We differentiate \( u \) with respect to \( x \), so \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2u} \). Therefore, \( dx = 2u \, du \). The limits of integration will change with this substitution: when \( x = 0 \), \( u = 0 \), and when \( x = \frac{\pi}{4} \), \( u = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2} \). Substituting these into the integral, we have:\[\int_{0}^{\frac{\sqrt{\pi}}{2}} \cos(u) \cdot (2u) \, du\]

Step 2: Set Up Integration by Parts

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). We choose: Let \( v = \cos(u) \), so \( dv = -\sin(u) \, du \), and let \( w = u \), so \( dw = du \). Thus, by integration by parts:\[\int u \, \cos(u) \, du = u \cdot \sin(u) - \int \sin(u) \, du\]

Step 3: Perform the Remaining Integration

Now integrate \( \int \sin(u) \, du \), which is \(-\cos(u)\). Substitute this back into the previous result:\[\int u \, \cos(u) \, du = u \sin(u) + \cos(u) + C\] Replace \( C \) with the proper limits of integration from \( 0 \) to \( \frac{\sqrt{\pi}}{2} \):\[(u \sin(u) + \cos(u)) \Bigg|_{0}^{\frac{\sqrt{\pi}}{2}}\]

Step 4: Evaluate the Integral

Substitute in the upper limit \( \frac{\sqrt{\pi}}{2} \):First term: \( \frac{\sqrt{\pi}}{2} \sin\left(\frac{\sqrt{\pi}}{2}\right) \) Second term: \( + \cos\left(\frac{\sqrt{\pi}}{2}\right) \)Now substitute the lower limit \( 0 \):First term: \( 0 \cdot \sin(0) = 0 \)Second term: \( + \cos(0) = 1 \)Thus the definite integral is:\[\frac{\sqrt{\pi}}{2} \sin\left(\frac{\sqrt{\pi}}{2}\right) + \cos\left(\frac{\sqrt{\pi}}{2}\right) - 1\]

Key Concepts

Substitution Method in Integration

The substitution method is a crucial technique in calculus that simplifies integrations by applying a change of variable. In the given problem, we started by choosing the substitution \( u = \sqrt{x} \). This is because substituting something that makes the expression simpler helps ease the integration process. As a rule of thumb:

• Identify a part of the integral that corresponds to a function (inside another function is a good place to start).
• Express \( dx \) in terms of the new variable \( du \).
• Calculate the corresponding changes in the limits of integration—originally from \( x \) to \( u \).

By computing \( dx = 2u \, du \), and changing the limits from \( 0 \) to \( \pi/4 \) into \( 0 \) to \( \sqrt{\pi}/2 \), the integral becomes significantly easier to handle. This conversion allows for easier further manipulations, such as the integration by parts used next.

Calculus Problems

Calculus problems often involve a series of steps to solve, using a combination of integration techniques. In the problem - integrating \( \int \cos \sqrt{x} \, dx \) - you saw the combination of substitution and integration by parts.

• Recognizing which method to use: Start with substitution if the integral includes compositions of functions.
• Strategy to break complex tasks: Use integration by parts for products of functions where one part is easier to integrate after differentiating the other.
• Sequential operations: As seen here, substitution simplifies an otherwise complex function, resetting the stage to employ effective methods like integration by parts.

Each of these steps demands attention to detail, but with practice, it becomes apparent how to identify which method serves the problem best.

Definite Integrals

Definite integrals compute the net area under a curve from one point to another. In the problem's solution, you worked to finding:\[\int_{0}^{\pi/4} \cos \sqrt{x} \, dx\]By doing substitutions and integration by parts, you put the definite integral into a simpler form, evaluating nearly every foundational element:

• Limits Substitution: Converting the original limits to the new ones after substitution ensures the integral remains definite.
• Computation: Upon applying integration by parts, the result showed explicit steps — integrating each resulting term separately and substituting back in the original step.
• Evaluating end values: Apply the new limits \( u = 0 \) and \( u = \sqrt{\pi}/2 \) – to get precise values after the integration formula is applied.

Thus, evaluating the definite integral gives you the result \( \frac{\sqrt{\pi}}{2} \sin\left( \frac{\sqrt{\pi}}{2} \right) + \cos\left( \frac{\sqrt{\pi}}{2} \right) - 1 \), which includes accounting for all parts—netting the area accurately represented by the integral.