Problem 55 - Calculus Early Transcendentals

Question

Additional integrals Evaluate the following integrals. $$\int_{0}^{\pi}(1-\cos 2 x)^{3 / 2} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $\int_{0}^{\pi}(1-\cos 2 x)^{3 / 2} d x$. Answer: $\frac{3\pi}{4}$.

Step 1: Using double angle formula for cosine

Double angle formula for cosine is: \(\cos 2x = 2\cos^2 x - 1\) Now, we will rewrite our integral as: $$\int_{0}^{\pi}(1-(2\cos^2 x - 1))^{3 / 2} d x$$

Step 2: Simplifying the integrand

Simplify the integrand: $$\int_{0}^{\pi}(1-2\cos^2 x + 1)^{3 / 2} d x ⇒ \int_{0}^{\pi}(2-2\cos^2 x)^{3 / 2} d x$$ Let \(u = \cos x\). Therefore, \(-2u^2 = 2-2\cos^2 x\). So the integrand becomes \((u^2)^{3/2} = u^3\). If \(u = \cos x\), then \(du = -\sin x dx\). We need to also change the boundaries of integration accordingly. When \(x=0\), \(u=\cos(0)=1\). When \(x=\pi\), \(u=\cos(\pi)=-1\). Our integral becomes: $$\int_{1}^{-1} -u^3 \left(-\frac{1}{\sin x}\right) du$$ or $$\int_{-1}^{1} u^3 \frac{1}{\sqrt{1-u^2}} du$$

Step 3: Substitution

We'll now make a substitution to further simplify the integrand. Let \(v = u^2\). Then \(dv = 2u du\). Now we update our integral: $$\frac{1}{2} \int_{0}^{1} \frac{v\sqrt{v}}{\sqrt{1-v}} dv$$

Step 4: Solving the integral

At this point, we can integrate our simplified function: $$\frac{1}{2} \int_{0}^{1} \frac{v\sqrt{v}}{\sqrt{1-v}} dv = \frac{1}{2} \int_{0}^{1} \frac{v^{3/2}}{(1-v)^{1/2}} dv$$ We solve the integral using a standard integral table entry: $$\frac{1}{2} \int_{0}^{1} \frac{v^{3/2}}{(1-v)^{1/2}} dv = \frac{1}{2} \left[B\left(\frac{5}{2},\frac{1}{2}\right)\right] ,$$ where \(B(x,y)\) is the Beta function. The Beta function is given by: $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ where \(\Gamma(z)\) is the Gamma function.

Step 5: Applying Beta function and Gamma function

Let's compute the Beta function for the integral using the formula: $$\frac{1}{2} \left[ \frac{\Gamma\left(\frac{5}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(3)} \right]$$ We're given that: $$\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$ So, $$\Gamma\left(\frac{5}{2}\right) = \frac{3}{2} \Gamma\left(\frac{3}{2}\right) = \frac{3}{2} \frac{1}{2} \Gamma\left(\frac{1}{2}\right) = \frac{3}{4} \sqrt{\pi}$$ $$\frac{1}{2} \left[ \frac{\frac{3}{4} \sqrt{\pi}\cdot\sqrt{\pi}}{\Gamma(3)} \right] = \frac{1}{2} \left[ \frac{3\pi}{2} \right] = \boxed{\frac{3\pi}{4}}$$ So the value of the integral is: $$\int_{0}^{\pi}(1-\cos 2 x)^{3 / 2} d x = \frac{3\pi}{4}$$

Key Concepts

Beta Function

The Beta function is a special integral expression, denoted as \( B(x, y) \). It is defined as:\[B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt\]The Beta function plays a critical role in calculus and analysis, especially in problems involving symmetrical properties and probability distributions.
It can transform complex expressions into simpler terms, making certain integrals easier to evaluate. One of its fascinating properties is its relationship with the Gamma function, defined by:\[B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\]This connection allows mathematicians to use known identities of the Gamma function to easily calculate the Beta function. In our original integral exercise, the Beta function helps simplify the complex left integral into a computable form.

Gamma Function

The Gamma function, denoted by \( \Gamma(z) \), extends the concept of factorial to complex numbers. It is defined as:\[\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt\]For natural numbers, it follows the simple relation: \( \Gamma(n) = (n-1)! \). For half-integers, like in our integral problem with \( \Gamma\left(\frac{1}{2}\right) \), special values include \( \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \).
The Gamma function offers a powerful tool for evaluating the Beta function and other integrals. In the solution above, we utilize the Gamma function to evaluate \( \Gamma\left(\frac{5}{2}\right) \) using its recursion properties.

Double Angle Formulas

Double angle formulas are mathematical identities used in trigonometry to express trigonometric functions of \(2x\) in terms of \(x\). A common double angle formula for cosine is:\[\cos(2x) = 2\cos^2(x) - 1\]These identities help simplify integrals or equations involving functions like \( \cos(2x) \). In the step by step solution of the original integral problem, the double angle formula is utilized to simplify the integrand.
This simplification allows for an easier manipulation of the integral to capitalize on other techniques, such as substitutions and changing the limits of integration.

Substitution in Integration

Substitution is a method used to simplify and evaluate integrals by changing the variable of integration. This technique involves replacing a part of the integrand with a new variable, which often turns the integral into a form that is easier to solve.
For instance, in our solution, the substitution \( u = \cos(x) \) simplifies the integral by changing the variable and modifying the integration limits from \([0, \pi] \) to \([-1, 1] \). After applying the substitution step, the integral becomes more straightforward to manage or solve using known integrals or further manipulations such as additional substitutions if necessary. Substitution is a key method in integral calculus for tackling more complex integrals.