Chapter 7: Problem 14
Evaluate the integral. $$\int_{0}^{\pi} \sin ^{2} t \cos ^{4} t d t$$
Short Answer
Expert verified
The integral evaluates to \( \frac{3\pi}{16} \).
Step by step solution
01
Use Trigonometric Identity
First, we need to simplify the integrand using a trigonometric identity. Recall that \( \sin^2 t = 1 - \cos^2 t \). However, for the product \( \sin^2 t \cdot \cos^4 t \), a better approach is to use the identity \( \sin^2 t = \frac{1 - \cos(2t)}{2} \). Thus, the integral becomes: \[ \int_{0}^{\pi} \left( \frac{1 - \cos(2t)}{2} \right) \cos^4 t \, dt. \]
02
Simplify the Integrand
Distribute the \( \frac{1}{2} \) into the integral: \[ \int_{0}^{\pi} \frac{1 - \cos(2t)}{2} \cos^4 t \, dt = \frac{1}{2} \int_{0}^{\pi} \cos^4 t \, dt - \frac{1}{2} \int_{0}^{\pi} \cos^4 t \cos(2t) \, dt. \]
03
Solve \( \int \cos^4 t \, dt \)
The first part of the integral is \( \frac{1}{2} \int_{0}^{\pi} \cos^4 t \, dt \). We can solve this by using the power-reduction formula: \[ \cos^4 t = \left( \cos^2 t \right)^2 = \left( \frac{1 + \cos(2t)}{2} \right)^2 = \frac{1 + 2\cos(2t) + \cos^2 (2t)}{4}. \]Now, integrate each term separately over the interval \([0, \pi]\).
04
Integrate Each Term
For \( \int_{0}^{\pi} 1 \, dt \), we have \( [t]_{0}^{\pi} = \pi \).For \( \int_{0}^{\pi} 2\cos(2t) \, dt \), use the substitution \( u = 2t \), so \( du = 2 \, dt \), leading to \( \int \cos u \, du \), which gives zero over the symmetric interval \([0, 2\pi]\). For \( \int_{0}^{\pi} \cos^2(2t) \, dt \), use the power-reduction formula to get \( \frac{1}{2} \int_{0}^{\pi} (1 + \cos(4t)) \, dt \).Evaluating this gives a total of \( \frac{3\pi}{8} \) for the first integral.
05
Simplify and Combine Results
The result of the integral part of \( \cos^4 t \) portion is effectively \( \frac{3\pi}{8} \, \frac{1}{2} = \frac{3\pi}{16} \).The second part of the original integral \( \int \cos^4 t \cos(2t) \, dt \) would result in zero over \([0, \pi]\) due to symmetry and periodic properties of objecting functions in odd multiples of their respective periods on \([0, \pi]\). Hence, the contribution from the complicated expression is zero.
06
Final Result
Combining all results, the value of the original integral is determined entirely by the simplified section of \( \cos^4 t \), yielding the answer:\[ \boxed{\frac{3\pi}{16}}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Identities
In this exercise, trigonometric identities play a crucial role in simplifying the original integral. Trigonometric identities allow us to express trigonometric functions in different, sometimes simpler, forms. One of the fundamental identities utilized here is \( \sin^2 t = \frac{1 - \cos(2t)}{2} \). This identity helps transform the product \( \sin^2 t \cdot \cos^4 t \) into a more manageable form for integration.
This identity stems from the double-angle identities, where \( \cos(2t) \) is related to \( \sin^2 t \) and \( \cos^2 t \). By using such identities, you can effectively reduce the complexity of expressions that contain powers of sine and cosine.
This identity stems from the double-angle identities, where \( \cos(2t) \) is related to \( \sin^2 t \) and \( \cos^2 t \). By using such identities, you can effectively reduce the complexity of expressions that contain powers of sine and cosine.
- Key Identity: \( \sin^2 t = \frac{1 - \cos(2t)}{2} \)
- Transformation: Simplifies \( \sin^2 t \cdot \cos^4 t \) into a form that can be integrated more easily.
Power-Reduction Formula
The power-reduction formula is another vital tool in this problem. When faced with higher powers of trigonometric functions, the power-reduction formulas simplify these powers into linear forms of trigonometric functions. For cosine, crucial formulas are derived like \( \cos^2 t = \frac{1 + \cos(2t)}{2} \), which help rewrite expressions involving powers of \( \cos(t) \) or \( \sin(t) \).
In this exercise, after rewriting \( \cos^4 t \), you apply the formula to break it down further:
In this exercise, after rewriting \( \cos^4 t \), you apply the formula to break it down further:
- Start with: \( \cos^4 t = (\cos^2 t)^2 \)
- Apply the power-reduction: \( \cos^2 t = \frac{1 + \cos(2t)}{2} \)
- Resulting Expression: \( \frac{1 + 2\cos(2t) + \cos^2 (2t)}{4} \)
Definite Integrals
Definite integrals compute the total accumulation of a function within a specific interval, providing a numeric value as a result. The bounds of the interval, in this case from \( 0 \) to \( \pi \), define the region of interest. Calculating definite integrals involves evaluating the antiderivative of the function at these bounds and finding the difference.
For the integral \( \int_{0}^{\pi} \sin^2 t \cos^4 t \ dt \), the process includes:
For the integral \( \int_{0}^{\pi} \sin^2 t \cos^4 t \ dt \), the process includes:
- Breaking down the complex integrand using identities and power-reduction.
- Solving separate integrals for each simplified expression.
- Using properties of symmetry and periodicity to simplify calculations where applicable.
- Applying the Fundamental Theorem of Calculus: \( F(b) - F(a) \)