Question

Show that (i) \(\int_{-a}^{a} \varphi\left(x^{2}\right) d x=2 \int_{0}^{a} \varphi\left(x^{2}\right) d x\) \(\int_{-a}^{a} x \varphi\left(x^{2}\right) d x=0\) (ii) \(\int_{0}^{\pi / 2} \varphi(\cos x) d x=\int_{0}^{\pi / 2} \varphi(\sin x) d x\) \(\quad=\frac{1}{2} \int_{0}^{\pi} \varphi(\sin x) d x\) (iii) \(\int_{0}^{m \pi} \phi\left(\cos ^{2} x\right) d x=m \int_{0}^{\pi} \phi\left(\cos ^{2} x\right) d x\),

Short answer

In this exercise, we proved the given integration formulas using properties of definite integrals, appropriate substitutions, and trigonometric identities. For part (i), we used the properties of even and odd functions to establish the integral formulas. In part (ii), we used trigonometric substitutions to prove the given integral equation. Finally, in part (iii), we used the periodic property of the cosine function to establish the given relationship involving the integral of the cosine function squared.

Step-by-step solution

Part (i) Definite Integral from -a to a

For the first equation, we need to show that: \(\int_{-a}^{a} \varphi\left(x^{2}\right) d x=2 \int_{0}^{a} \varphi\left(x^{2}\right) d x\) Step 1: Apply the property of even functions - Using the property that the integral of a function between symmetric limits can be expressed as: \(\int_{-a}^a f(x) dx = 2 \int_{0}^a f(x) dx\) where \(f(x)\) is an even function. Step 2: Identify \(\varphi\left(x^{2}\right)\) as an even function - An even function is defined as \(f(-x) = f(x)\). Since \(\varphi\left(x^{2}\right) = \varphi\left((-x)^{2}\right) = \varphi\left(x^{2}\right)\), \(\varphi\left(x^{2}\right)\) is an even function. Step 3: Apply the even function property to the given equation - Now we can use the result from step 1 to rewrite the integral as: \(\int_{-a}^{a} \varphi\left(x^{2}\right) d x=2 \int_{0}^{a} \varphi\left(x^{2}\right) d x\) For the second equation, we need to show that: \(\int_{-a}^{a} x \varphi\left(x^{2}\right) d x=0\) Step 1: Apply the property of odd functions - Using the property that the integral of an odd function between symmetric limits is zero: \(\int_{-a}^a f(x) dx = 0\) where \(f(x)\) is an odd function. Step 2: Identify \(x\varphi\left(x^{2}\right)\) as an odd function - An odd function is defined as \(f(-x) = -f(x)\). Since \(x\varphi\left(x^{2}\right) = -(-x)\varphi\left((-x)^{2}\right) = -x\varphi\left(x^{2}\right)\), \(x\varphi\left(x^{2}\right)\) is an odd function. Step 3: Apply the odd function property to the given equation - Now we can use the result from step 1 to rewrite the integral as: \(\int_{-a}^{a} x \varphi\left(x^{2}\right) d x=0\)

Part (ii) Definite Integral with Trigonometric Functions

We need to show that: \(\int_{0}^{\pi / 2} \varphi(\cos x) d x=\int_{0}^{\pi / 2} \varphi(\sin x) d x = \frac{1}{2} \int_{0}^{\pi} \varphi(\sin x) d x\) Step 1: Substitute \(x\) with \(\pi/2 - x\) in the first equation - The first integral will change to its equivalent expression: \(\int_{0}^{\pi / 2} \varphi(\sin (\pi/2 - x)) dx\). Since \(\cos(\pi/2 - x) = \sin x\), we can rewrite it as: \(\int_{0}^{\pi / 2} \varphi(\sin x) d x\). Step 2: Identity between the first two integrals - Now, the first two equations are identical, so we have: \(\int_{0}^{\pi / 2} \varphi(\cos x) d x = \int_{0}^{\pi / 2} \varphi(\sin x) d x\). Step 3: Transformation for the third integral - Using the change of variables, we will substitute \(u = \pi - x\) in the third equation. Since \(du = -dx\), we have: \(\frac{1}{2} \int_{0}^{\pi} \varphi(\sin x) d x = -\frac{1}{2} \int_\pi^0 \varphi(\sin u) du\). Step 4: Change the bounds of the integral - Now, we change the bounds of the integral: \(-\frac{1}{2} \int_\pi^0 \varphi(\sin u) du = \frac{1}{2} \int_0^\pi \varphi(\sin u) du\). Step 5: Identity between the second and third integrals - Finally, we have: \(\int_{0}^{\pi / 2} \varphi(\sin x) d x = \frac{1}{2} \int_{0}^{\pi} \varphi(\sin x) d x\).

Part (iii) Definite Integral with Trigonometric Identities

In this part, we need to show that: \(\int_{0}^{m \pi} \phi\left(\cos ^{2} x\right) d x=m \int_{0}^{\pi} \phi\left(\cos ^{2} x\right) d x\) Step 1: Periodic property of \(\cos^{2}(x)\) - We know that \(\cos^{2}(x)\) is a periodic function with period \(\pi\). Therefore, \(\phi\left(\cos ^{2} x\right)\) has period \(\pi\) as well. Step 2: Express the integral as the sum of periodic functions - Since \(\phi\left(\cos ^{2} x\right)\) has period \(\pi\), we can express the integral as the sum of periodic functions: \(\int_{0}^{m \pi} \phi\left(\cos ^{2} x\right) d x = \int_{0}^{\pi} \phi\left(\cos ^{2} x\right) d x + \int_{\pi}^{2 \pi} \phi\left(\cos ^{2} x\right) d x + \dots + \int_{(m-1) \pi}^{m \pi} \phi\left(\cos ^{2} x\right) d x\). Step 3: Use the periodic property of \(\phi\left(\cos ^{2} x\right)\) - Since \(\phi\left(\cos ^{2} x\right)\) is periodic with period \(\pi\), we have: \(\int_{0}^{\pi} \phi\left(\cos ^{2} x\right) d x = \int_{\pi}^{2 \pi} \phi\left(\cos ^{2} x\right) d x = \dots = \int_{(m-1) \pi}^{m \pi} \phi\left(\cos ^{2} x\right) d x\). Step 4: Combine the integrals - Therefore, the integral from \(0\) to \(m\pi\) can be expressed as: \(\int_{0}^{m \pi} \phi\left(\cos ^{2} x\right) d x=m \int_{0}^{\pi} \phi\left(\cos ^{2} x\right) d x\).