Question

$$\text {Evaluate the following integrals.}$$ $$\int_{0}^{\sqrt{\pi / 2}} x \sin ^{3}\left(x^{2}\right) d x$$

Short answer

Question: Evaluate the definite integral $\int_{0}^{\sqrt{\pi / 2}} x \sin ^{3}\left(x^{2}\right) d x$. Answer: $\frac{1}{4}$.

Step-by-step solution

Identify the substitution

Let's substitute $$u = x^2$$. This will help to simplify the integral and make it easier to evaluate.

Differentiate with respect to x

To perform the substitution, we'll need the derivative of $$u$$ with respect to $$x.$$ Differentiate $$u = x^2$$ with respect to $$x$$ to get: $$\frac{d u}{d x}=2 x$$

Change the bounds of integration

Since we are changing variables, we must also change the limits of integration accordingly. The given integral has bounds $$0$$ and $$\sqrt{\pi/2}$$. For the substitution $$u = x^2$$, use these values of $$x$$ to find the corresponding values of $$u:$$ When $$x=0$$, $$u=0^2 = 0$$ When $$x=\sqrt{\pi/2}$$, $$u=\left(\sqrt{\pi/2}\right)^2 = \pi/2$$ Thus, the new bounds of integration are $$u=0$$ and $$u=\pi/2$$.

Rewrite the integral and perform the substitution

Now, we can rewrite the original integral in terms of $$u$$ by substituting $$u = x^2$$ and dividing by the derivative found in Step 2, $$\frac{d u}{d x}=2 x$$: $$\int_{0}^{\pi/2} x \sin ^3(u) \cdot \frac{1}{2x} d u$$ Notice that $$x$$ in the numerator and denominator cancel each other. Our substituted integral becomes: $$\frac{1}{2}\int_{0}^{\pi/2} \sin ^3(u) d u$$

Evaluate the integral

Now, let's evaluate the integral $$\frac{1}{2}\int_{0}^{\pi/2} \sin ^3(u) d u$$. To do this, we will use the power-reduction formula for sine: $$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$ First, write $$\sin^3(u)$ as $$\sin(u)\sin^2(u)$$ and substitute the power-reduction formula: $$\frac{1}{2}\int_{0}^{\pi/2} \sin (u) \cdot \frac{1 - \cos(2u)}{2} d u$$ Distribute $$\sin(u)$$: $$\frac{1}{4}\int_{0}^{\pi/2} (\sin(u) - \sin(u)\cos(2u)) d u$$ Now, evaluate the integral of each term separately: $$\frac{1}{4}\left[\int_{0}^{\pi/2} \sin(u) d u - \int_{0}^{\pi/2} \sin(u)\cos(2u) d u\right]$$ The first integral can be computed directly. For the second integral, we make a substitution, $$v = 2u$$: $$\frac{d v}{d u} = 2 \Rightarrow d u = \frac{1}{2} d v$$ When $$u=0$$, $$v=0$$ When $$u=\pi/2$$, $$v=\pi$$ This gives us: $$\frac{1}{4}\left[\int_{0}^{\pi/2} \sin(u) d u - \frac{1}{2}\int_{0}^{\pi} \frac{\sin(v/2)}{2}\cos(v) d v\right]$$ Now, we can integrate each term: $$\frac{1}{4}\left[-\cos(u)\Big|_0^{\pi/2} - \frac{1}{4}\left(\frac{\sin^2(v)}{2}\right)\Big|_0^{\pi}\right]$$

Evaluate the bounds

Finally, evaluate each term at the corresponding bounds: $$\frac{1}{4}\left[-\left(\cos\left(\frac{\pi}{2}\right) - \cos(0)\right) - \frac{1}{4}\left(\left(\frac{\sin^2\left(\pi\right)}{2}\right)-\left(\frac{\sin^2(0)}{2}\right)\right)\right]$$ This simplifies to: $$\frac{1}{4}\left[-(-1) - \frac{1}{4}(0)\right] = \frac{1}{4}$$ Thus, the value of the integral $$\int_{0}^{\sqrt{\pi / 2}} x \sin ^{3}\left(x^{2}\right) d x$$ is $$\frac{1}{4}$$.

Key concepts

Definite Integrals

In integral calculus, definite integrals are used to calculate the area under a curve between two points on the x-axis. This involves finding the antiderivative of a function over a given interval. When you compute a definite integral, you are essentially summing up an infinite number of infinitely small rectangles under a curve between the limits of integration.

• The integral sign \( \int \) denotes the operation of integration.
• The numbers at the top and bottom of the integral sign are the limits of integration.
• The bottom number is the lower limit, and the top number is the upper limit.

The definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x)\, dx \) and represents the net area under the curve. It’s crucial to have the correct limits to ensure the integral is computed over the desired interval. This becomes especially important when the problem involves a substitution, as the limits must be changed according to the new variable that’s introduced through substitution.

Trigonometric Integrals

Trigonometric integrals are integrals involving trigonometric functions like \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \). These integrals often require special techniques because trigonometric functions have specific properties that make integration more challenging than for polynomial functions.
One common technique is using trigonometric identities to simplify the integral before solving. For example, power-reduction formulas can be used to simplify integrals involving higher powers of sine or cosine, as was done in this solution.

• The power-reduction formula for \( \sin^2(x) \) is \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \).
• Using this formula, you can break down a complicated trigonometric integral into simpler parts.
• Each part can then be integrated separately, often involving another substitution to simplify the process.

This is especially useful because it helps transform an integral into a more recognizable form that can be evaluated directly or with further substitutions. Understanding these identities and when to use them is key in solving trigonometric integrals effectively.

Substitution Method

The substitution method, also known as "u-substitution," is an essential tool in integral calculus. It is particularly useful for simplifying integrals that are not easily solvable in their original form. The idea is to transform the integral into a simpler form by changing the variable.
Here’s how substitution works:

• Identify a part of the integral that can be considered as \( u \), such that the rest of the expression is the derivative of \( u \).
• Replace the original variable and differential with \( u \) and \( du \), making sure to adjust the differential according to the chain rule.
• Change the limits of the integral if it is definite, like in this example, by substituting the original limits into the expression for \( u \).

This method transforms a complex integral into a simpler one, often turning a trigonometric or polynomial expression into a basic form. In this problem, \( x^2 \) was substituted as \( u \), converting the complex integral into one involving \( u \) and \( du \) that was easier to handle with further algebraic manipulation and trigonometric identities.