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Magazine Chapter 8: Problem 29
Evaluate the integrals $$\int_{5 \pi / 6}^{\pi} \frac{\cos ^{4} x}{\sqrt{1-\sin x}} d x$$
Short Answer
Expert verified
The integral evaluates to approximately 0.383.
Step by step solution
01
Identify the Substitution
To evaluate this integral, we will use a substitution to simplify the integrand. Notice that the term under the square root is of the form \(1 - \sin x\). A common substitution for this type of expression is \(u = 1 - \sin x\) because it will simplify the integral by eliminating the square root.
02
Determine du and Change Limits
After choosing \(u = 1 - \sin x\), differentiate to find \(du\). We have \(\frac{du}{dx} = -\cos x\), which gives \(du = -\cos x \, dx\). Additionally, change the limits of integration: when \(x = 5\pi/6\), \(u = 1 - \sin(5\pi/6) = 1 - \frac{1}{2} = \frac{1}{2}\); and when \(x = \pi\), \(u = 1 - \sin(\pi) = 1 - 0 = 1\).
03
Simplify the Integral
Substitute \(u = 1 - \sin x\) and \(dx = -\frac{du}{\cos x}\) into the integral. The integral becomes:\[\int_{1/2}^{1} \frac{(1-u)^2}{\sqrt{u}} (-du) = \int_{1/2}^{1} (1 - 2u + u^2) u^{-1/2} (-du)\]Change the sign of the integral to simplify further:
04
Integrate Term by Term
Distribute the \(u^{-1/2}\) and then integrate each term separately:\[-\int_{1/2}^{1} u^{-1/2} \, du + 2 \int_{1/2}^{1} u^{1/2} \, du - \int_{1/2}^{1} u^{3/2} \, du\]Integrate each term using power rules:\[- (u^{1/2}) + 2 \left(\frac{2}{3} u^{3/2} \right) - \left( \frac{2}{5} u^{5/2} \right)\] evaluated from \(1/2\) to \(1\).
05
Evaluate the Definite Integral
Substitute the limits into the antiderivative and compute:- For the first term: \(-(u^{1/2})) = -1 + \sqrt{1/2}\).- For the second term: \(\frac{4}{3}(u^{3/2}) = \frac{4}{3} \times 1 - \frac{4}{3}(\frac{1}{2})^{3/2}\).- For the third term: \(-\frac{2}{5}(u^{5/2}) = -\frac{2}{5}(1) + \frac{2}{5}(\frac{1}{2})^{5/2}\).Combine these results.
06
Simplify the Result
Combine the evaluated parts to get a simplified expression. After substitutions, the result simplifies to a single number:\[-1 + \sqrt{\frac{1}{2}} + \frac{4}{3} \left( 1 - \frac{1}{8} \right) - \frac{2}{5} \left( 1 - \frac{1}{\sqrt{32}} \right)\].Re-evaluate and combine all terms to finalize the solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a central concept in calculus, giving us the means to calculate the net area under a curve bounded by two vertical lines on the x-axis. They differ from indefinite integrals because they calculate the precise quantified value over an interval, whereas indefinite integrals represent a family of functions.
- The notation for a definite integral is given by \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the endpoints of the interval.
- In a definite integral, you evaluate the result from the lower bound \( a \) to the upper bound \( b \), using the antiderivative of \( f(x) \).
Trigonometric Substitution
Trigonometric substitution is a powerful technique used to simplify integrals involving square roots of trigonometric functions. When certain expressions transform into a more manageable form through trigonometric identities, integration becomes less complicated.
- This method often involves using identities such as \( \sin^2 x + \cos^2 x = 1 \) to replace parts of the integrand.
- In the given problem, since \( 1 - \sin x \) is present under the square root, a substitution such as \( u = 1 - \sin x \) helps eliminate the complexity caused by the square root.
Integration by Substitution
Integration by substitution, often referred to as \( u \)-substitution, is akin to the reverse of the chain rule for differentiation. It simplifies finding antiderivatives by changing variables to transform the integral into an easier form.
- The trick involves selecting a substitution \( u = g(x) \) such that the derivative \( du = g'(x) \, dx \) can replace part of the integrand.
- Don't forget to change the limits of integration according to the new variable \( u \) when dealing with definite integrals.
