Problem 27 - Biocalculus Calculus for the Life Sciences

Question

(a) If \(n \geqslant 2\) is an integer, show that $$\int \sin ^{n} x d x=-\frac{1}{n} \cos x \sin ^{n-1} x+\frac{n-1}{n} \int \sin ^{n-2} x d x$$ This is called a reduction formula because the exponent \(n\) has been reduced to \(n-1\) and \(n-2 .\) (b) Use the reduction formula in part (a) to show that $$\int \sin ^{2} x d x=\frac{x}{2}-\frac{\sin 2 x}{4}+C$$ (c) Use parts (a) and (b) to evaluate \(\int \sin ^{4} x d x\) .

Step-by-Step Solution

Answer

\(\int \sin^n x dx = -\frac{1}{n}\cos x \sin^{n-1} x + \frac{n-1}{n}\int \sin^{n-2} x dx\); \(\int \sin^2 x dx = \frac{x}{2} - \frac{\sin 2x}{4} + C\); \(\int \sin^4 x dx = -\frac{1}{4}\cos x \sin^3 x + \frac{3x}{8} - \frac{3\sin 2x}{16} + C\).

Step 1: Understanding the Reduction Formula

The goal is to derive the reduction formula given in the problem:\[ \int \sin^n x \, dx = -\frac{1}{n} \cos x \sin^{n-1} x + \frac{n-1}{n} \int \sin^{n-2} x \, dx \]This formula helps reduce the power of \( \sin x \) by breaking it down into expressions with lower powers.

Step 2: Applying Integration by Parts

To derive the reduction formula, apply integration by parts to \( \int \sin^n x \, dx \). Let:- \( u = \sin^{n-1} x \) and \( dv = \sin x \, dx \).Then, \( du = (n-1) \sin^{n-2}x \cos x \, dx \) and \( v = -\cos x \).Apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \).

Step 3: Deriving the Reduction Formula

Compute \( uv - \int v \, du \):\[ uv = -\cos x \sin^{n-1} x \]\[ \int v \, du = - (n-1) \int \cos^2 x \sin^{n-2} x \, dx \]Since \( \cos^2 x = 1 - \sin^2 x \), substitute this in:\[ - (n-1) \int \sin^{n-2} x (1 - \sin^2 x) \, dx = - (n-1) \left( \int \sin^{n-2} x \, dx - \int \sin^n x \, dx \right) \]Simplify to obtain the reduction formula.

Step 4: Proving the Reduction Formula

After simplifying, we have:\[ \int \sin^n x \, dx = -\cos x \sin^{n-1} x + (n-1) \int \sin^{n-2} x \, dx \]Divide through by \( n \) to get the desired formula:\[ \int \sin^n x \, dx = -\frac{1}{n} \cos x \sin^{n-1} x + \frac{n-1}{n} \int \sin^{n-2} x \, dx \]

Step 5: Evaluating \( \int \sin^2 x \, dx \)

Using the reduction formula with \( n = 2 \):\[ \int \sin^2 x \, dx = -\frac{1}{2} \cos x \sin^{1} x + \frac{1}{2} \int \sin^0 x \, dx \]This simplifies to:\[ -\frac{1}{2} \cos x \sin x + \frac{1}{2} x \]Recognizing \( -\frac{1}{2} \cos x \sin x = -\frac{\sin 2x}{4} \), we get:\[ \int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C \]

Step 6: Evaluating \( \int \sin^4 x \, dx \) Using Previous Results

Use the reduction formula for \( n = 4 \):\[ \int \sin^4 x \, dx = -\frac{1}{4} \cos x \sin^3 x + \frac{3}{4} \int \sin^2 x \, dx \]Substitute \( \int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C \) into the formula:\[ \int \sin^4 x \, dx = -\frac{1}{4} \cos x \sin^3 x + \frac{3}{4} \left( \frac{x}{2} - \frac{\sin 2x}{4} + C \right) \]

Step 7: Simplifying the Solution for \( \int \sin^4 x \, dx \)

Simplify the expression:\[ \int \sin^4 x \, dx = -\frac{1}{4} \cos x \sin^3 x + \frac{3x}{8} - \frac{3\sin 2x}{16} + C \]

Key Concepts

Integration by Parts

To grasp the reduction formula for trigonometric integrals, one needs a good handle on integration by parts. This technique is a cornerstone in calculus for integrating products of functions. It's an extension of the product rule for differentiation and is expressed by the formula:

• \( \int u \, dv = uv - \int v \, du \).

The idea is to wisely choose the functions \( u \) and \( dv \) such that the resulting integral \( \int v \, du \) is simpler to solve. By applying this process repeatedly, more complex integration tasks become manageable.
In our original exercise, the integration of \( \sin^n(x) \) was tackled by choosing \( u = \sin^{n-1}(x) \) and \( dv = \sin(x) \, dx \). The derivative \( du \) and integral \( v \) are then computed, leading to new expression forms that simplify the integral step-by-step.

Trigonometric Integrals

Trigonometric integrals often involve functions like \( \sin(x) \), \( \cos(x) \), and powers thereof. Solving these involves specific strategies, like using trigonometric identities or applying reduction formulas. The reduction formula given in the exercise decreases the power of the sine function, making repeated integration simpler.
A good approach in tackling trigonometric integrals is to:

• Utilize trigonometric identities to rewrite integrals.
• Consider reduction formulas for integration when dealing with higher powers.
• Apply symmetry properties of trigonometric functions, such as periodicity, to simplify integration limits.

These strategies help reduce complex integrations into more approachable forms.

Sin Function

The sine function, \( \sin(x) \), is one of the fundamental trigonometric functions, representing oscillations and waves. Mathematically, it's defined as the y-coordinate of a point on the unit circle corresponding to an angle \( x \).
When dealing with integrals of sine functions, understanding its properties such as periodicity (\( 2\pi \) radians) is critical. This periodicity can sometimes be leveraged to simplify the calculations of definite integrals.

• Another key aspect is the identity \( \cos^2(x) = 1 - \sin^2(x) \), often helpful in manipulating sine powers.

This was used in our problem to decompose powers of sine into more solvable integrals during the reduction process.

Calculus Problem Solving

In calculus, problem-solving often requires a structured approach. Problems like those involving trigonometric integrals can seem daunting, but breaking them into smaller parts makes them more digestible. Here’s a strategy for dealing with these problems:

• Carefully read and understand the problem statement.
• Identify the mathematical techniques required, such as integration by parts or reduction formulas.
• Break the problem down and solve it step by step, double-checking your intermediate results.
• Keep track of constants of integration when dealing with indefinite integrals.

For the problem at hand, using a systematic approach led to the stepwise derivation of a reduction formula and application to specific powers of sine. This structured method is vital in ensuring clarity and accuracy in solving complex calculus problems.