Question

$\int_{\pi/4}^{\pi/2} 2\sin^{-4} x dx = \_\_\_\_\_\_.$
Note: The initial OCR showed "$23.4 \frac{/2}{/4}$". The "4" was a misread coefficient. The mathematical evaluation of the options indicates a coefficient of 2 is present in the intended question.

• A. 8/3
• B. -8/3
• C. 2/3
• D. -2/3

Hint:

For any integral involving $\sec^4(x)$ or $\csc^4(x)$, separating out a squared term to serve as the derivative ($du$) for a $u = \tan(x)$ or $u = \cot(x)$ substitution is the standard, foolproof method.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We must evaluate a definite integral of $\sin^{-4}(x)$, which is algebraically equivalent to $\csc^4(x)$.

Step 2: Key Formula or Approach:
To integrate even powers of cosecant, factor out a $\csc^2(x)$ and convert the remaining $\csc^2(x)$ terms into $\cot^2(x)$ using the Pythagorean identity $1 + \cot^2(x) = \csc^2(x)$. Then, apply $u$-substitution.

Step 3: Detailed Explanation:
Let's evaluate the base integral $I = \int_{\pi/4}^{\pi/2} \csc^4 x \, dx$.
Rewrite the integrand:
$$I = \int_{\pi/4}^{\pi/2} \csc^2 x \cdot \csc^2 x \, dx$$
$$I = \int_{\pi/4}^{\pi/2} (1 + \cot^2 x) \csc^2 x \, dx$$
Let $u = \cot x$. Differentiating gives $du = -\csc^2 x \, dx \implies -du = \csc^2 x \, dx$.
Change the bounds of integration:
When $x = \pi/4$, $u = \cot(\pi/4) = 1$.
When $x = \pi/2$, $u = \cot(\pi/2) = 0$.
Substitute into the integral:
$$I = \int_{1}^{0} (1 + u^2) (-du) = \int_{0}^{1} (1 + u^2) \, du$$
Integrate with respect to $u$:
$$I = \left[ u + \frac{u^3}{3} \right]_0^1 = \left( 1 + \frac{1}{3} \right) - (0) = \frac{4}{3}$$
The base integral evaluates to $4/3$.
The integral possesses a constant multiplier of 2, making the final value:
$$2 \times I = 2 \times \frac{4}{3} = \frac{8}{3}$$

Step 4: Final Answer:
The result is 8/3, which matches option (a).