Problem 31 - Calculus for Biology and Medicine

Question

Evaluating the integral $$ \int \cos ^{2} x d x $$ First, write $$ \cos ^{2} x=(\cos x)(\cos x) $$ and integrate by parts to show that $$ \int \cos ^{2} x d x=\sin x \cos x+\int \sin ^{2} x d x $$ Then, use \(\sin ^{2} x+\cos ^{2} x=1\) to replace \(\sin ^{2} x\) in the integral on the right-hand side, and complete the integration of \(\int \cos ^{2} x d x\).

Step-by-Step Solution

Answer

\( \int \cos^2 x \, dx = \frac{1}{2}(\sin x \cos x + x) + C \).

Step 1: Set Up Integration by Parts

To integrate by parts, remember the formula \( \int u \ dv = uv - \int v \ du \). Use \( u = \cos x \) and \( dv = \cos x \ dx \), giving us \( du = -\sin x \ dx \) and \( v = \int \cos x\, dx = \sin x \).

Step 2: Apply Integration by Parts Formula

Substitute the chosen \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \( \int \cos^2 x \, dx = \cos x \cdot \sin x - \int \sin x \cdot (-\sin x) \, dx \). Simplify this to: \( \int \cos^2 x \, dx = \sin x \cos x + \int \sin^2 x \, dx \).

Step 3: Substitute Using Trigonometric Identity

Use the identity \( \sin^2 x + \cos^2 x = 1 \) to express \( \sin^2 x \) as \( 1 - \cos^2 x \). Substitute this into the integral: \( \int \cos^2 x \, dx = \sin x \cos x + \int (1 - \cos^2 x) \, dx \).

Step 4: Break Down and Solve the Resulting Integral

Write the integral as two separate integrals: \( \int \cos^2 x \, dx = \sin x \cos x + \int 1 \, dx - \int \cos^2 x \, dx \). This simplifies to: \( 2\int \cos^2 x \, dx = \sin x \cos x + x \).

Step 5: Solve for the Original Integral

Isolate \( \int \cos^2 x \, dx \) by dividing both sides of the equation by 2: \( \int \cos^2 x \, dx = \frac{1}{2}(\sin x \cos x + x) + C \), where \( C \) is the constant of integration.

Key Concepts

Integration by Parts

Integration by parts is a clever technique in integral calculus. It is especially helpful when we're dealing with the product of two functions. The formula behind this technique is inspired by the product rule from differentiation:
\[ \int u \ dv = uv - \int v \ du \]
Here's the step-by-step simple breakdown to make it more digestible:

• Choose functions for \( u \) and \( dv \) such that \( u \) is easy to differentiate and \( dv \) is easy to integrate.
• Once you select \( u \) and \( dv \), compute \( du \) by differentiating \( u \), and \( v \) by integrating \( dv \).
• Substitute these into the integration by parts formula to simplify the problem.

When evaluating \( \int \cos^2 x \ dx \), if we choose \( u = \cos x \) and \( dv = \cos x \ dx \), the solution naturally unfolds into manageable parts leading us to easier integrals that we can solve.

Trigonometric Identities

Trigonometric identities are like the secret passages in calculus that make problems more approachable. These identities can transform expressions, simplify calculations, and in our case, solve integrals.

• One fundamental identity is \( \sin^2 x + \cos^2 x = 1 \). It helps us express one trigonometric square function in terms of another.
• For our exercise, we use this identity to replace \( \sin^2 x \) with \( 1 - \cos^2 x \).

This simplification is key as it allows us to rewrite the original integral into forms that are more straightforward to solve. If you can remember this key identity, it frequently simplifies what looks like a tough calculus problem into something much easier!

Definite Integrals

Our discussion focuses on indefinite integrals, but definite integrals are closely related and are a crucial piece of calculus. They represent the actual area under a curve between two specific points on the x-axis.

• Definite integrals are written as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• The Fundamental Theorem of Calculus connects differentiation and integration, easing the calculation of definite integrals using antiderivatives.

While not directly solving for a definite integral in this exercise, understanding indefinite integrals thoroughly is the first step. Mastery of these concepts will make dealing with the bounds introduced in definite integrals a breeze later on.