Evaluate the following integral.
Integral from 0 to StartRoot StartFraction pi Over 2 EndFraction EndRoot 7 x sine cubed left parenthesis x squared right parenthesis dx
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Part 1
Integral from 0 to StartRoot StartFraction pi Over 2 EndFraction EndRoot 7 x sine cubed left parenthesis x squared right parenthesis dx
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enter your response here ​(Type an integer or a simplified​ fraction.)

We are tasked with evaluating the following definite integral:

\[
\int_0^{\sqrt{\frac{\pi}{2}}} 7x \sin^3(x^2) \, dx
\]

### Step 1: Substitution
To simplify this integral, we can use the substitution \( u = x^2 \), since \( x^2 \) appears inside the sine function.

- \( du = 2x \, dx \), so \( \frac{du}{2} = x \, dx \).

Now, we change the limits of integration:
- When \( x = 0 \), \( u = 0^2 = 0 \).
- When \( x = \sqrt{\frac{\pi}{2}} \), \( u = \left(\sqrt{\frac{\pi}{2}}\right)^2 = \frac{\pi}{2} \).

Thus, the integral becomes:

\[
\int_0^{\frac{\pi}{2}} 7 \sin^3(u) \cdot \frac{du}{2}
\]

Simplifying:

\[
\frac{7}{2} \int_0^{\frac{\pi}{2}} \sin^3(u) \, du
\]

### Step 2: Use a reduction formula for \(\sin^3(u)\)
The next step is to simplify \( \sin^3(u) \). We can use the identity:

\[
\sin^3(u) = \sin(u)(1 - \cos^2(u))
\]

Thus, the integral becomes:

\[
\frac{7}{2} \int_0^{\frac{\pi}{2}} \sin(u) (1 - \cos^2(u)) \, du
\]

Now, we can split this into two separate integrals:

\[
\frac{7}{2} \left( \int_0^{\frac{\pi}{2}} \sin(u) \, du - \int_0^{\frac{\pi}{2}} \sin(u) \cos^2(u) \, du \right)
\]

### Step 3: Solve the integrals

1. The first integral is straightforward:

\[
\int_0^{\frac{\pi}{2}} \sin(u) \, du = \left[ -\cos(u) \right]_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1
\]

2. The second integral requires substitution. Let \( v = \cos(u) \), so \( dv = -\sin(u) \, du \). When \( u = 0 \), \( v = 1 \), and when \( u = \frac{\pi}{2} \), \( v = 0 \). Thus, the second integral becomes:

\[
-\int_1^0 v^2 \, (-dv) = \int_0^1 v^2 \, dv
\]

This is a standard integral:

\[
\int_0^1 v^2 \, dv = \left[ \frac{v^3}{3} \right]_0^1 = \frac{1}{3}
\]

### Step 4: Combine the results

Now, we substitute the results of the two integrals back into the expression:

\[
\frac{7}{2} \left( 1 - \frac{1}{3} \right) = \frac{7}{2} \left( \frac{2}{3} \right) = \frac{7 \times 2}{6} = \frac{14}{6} = \frac{7}{3}
\]

Thus, the value of the integral is:

\[
\boxed{\frac{7}{3}}
\]

Would you like further details or have any other questions?

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**Related Questions:**
1. How would the integral change if the limits were altered?
2. Can this integral be solved using a different method, like integration by parts?
3. What if the integrand was \( \sin^n(x^2) \)? How does the complexity change?
4. How does substitution help in solving integrals of this type?
5. Could this integral be evaluated using numerical methods?

**Tip:** Always consider trigonometric identities or substitutions when simplifying integrals involving powers of trigonometric functions.

Evaluate the following integral: Integral from 0 to sqrt(pi/2) of 7x sin^3(x^2) dx

Learn how to evaluate the definite integral from 0 to sqrt(pi/2) of 7x sin^3(x^2) dx using substitution and trigonometric identities. Follow a detailed step-by-step explanation, covering key concepts like the substitution method and reduction formulas.

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