The integral provided in the image is:

$\int_{\pi/4}^{\pi/2} \frac{dx}{\sin^4(x)}.$

Let's solve this step by step:

Step 1: Rewrite the integral

We know that: $\frac{1}{\sin^4(x)} = \csc^4(x),$ so the integral becomes: $\int_{\pi/4}^{\pi/2} \csc^4(x) \, dx.$

Step 2: Use the reduction formula for $\csc^n(x)$

The reduction formula for $\int \csc^n(x) \, dx$ (when $n > 2$) is: $\int \csc^n(x) \, dx = -\frac{\csc^{n-2}(x) \cot(x)}{n-1} + \frac{n-2}{n-1} \int \csc^{n-2}(x) \, dx.$

Here, $n = 4$. Applying the formula: $\int \csc^4(x) \, dx = -\frac{\csc^2(x) \cot(x)}{3} + \frac{2}{3} \int \csc^2(x) \, dx.$

Step 3: Solve $\int \csc^2(x) \, dx$

The integral of $\csc^2(x)$ is well-known: $\int \csc^2(x) \, dx = -\cot(x).$

Substitute this into the formula: $\int \csc^4(x) \, dx = -\frac{\csc^2(x) \cot(x)}{3} - \frac{2}{3} \cot(x).$

Step 4: Evaluate the definite integral

Now we evaluate $\int_{\pi/4}^{\pi/2} \csc^4(x) \, dx$: $\left[ -\frac{\csc^2(x) \cot(x)}{3} - \frac{2}{3} \cot(x) \right]_{\pi/4}^{\pi/2}.$

Evaluate at $x = \pi/2$:

$\csc(\pi/2) = 1$ and $\cot(\pi/2) = 0$. Substituting these: $-\frac{\csc^2(\pi/2) \cot(\pi/2)}{3} - \frac{2}{3} \cot(\pi/2) = 0.$

Evaluate at $x = \pi/4$:

$\csc(\pi/4) = \sqrt{2}$ and $\cot(\pi/4) = 1$. Substituting these: $-\frac{\csc^2(\pi/4) \cot(\pi/4)}{3} - \frac{2}{3} \cot(\pi/4) = -\frac{(\sqrt{2})^2 (1)}{3} - \frac{2}{3}(1) = -\frac{2}{3} - \frac{2}{3} = -\frac{4}{3}.$

Thus, the result is: $0 - \left(-\frac{4}{3}\right) = \frac{4}{3}.$

Final Answer:

$\int_{\pi/4}^{\pi/2} \frac{dx}{\sin^4(x)} = \frac{4}{3}.$

Would you like further details on any step or related explanations?

Related Questions:

1. What is the derivation of the reduction formula for $\csc^n(x)$?
2. How do you evaluate integrals involving powers of $\sec(x)$ or $\csc(x)$?
3. Can we solve this integral using trigonometric identities instead of reduction formulas?
4. What is the significance of improper integrals involving trigonometric functions?
5. How does the value of $\pi/4$ and $\pi/2$ impact the trigonometric function evaluations?

Tip:

For integrals involving trigonometric powers, reduction formulas or substitutions like $\sin(x) = t$ can often simplify the computation!