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\( \left. \begin{array} { l } { \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \sin x ) ^ { \frac { 8 } { 3 } } ( \sec x ) ^ { \frac { 1 } { 2 } } d x } \\ { \pi } \\ { \int _ { 0 } ^ { \pi } \sin x ) ^ { \frac { 7 } { 2 } } \cos ^ { \frac { 3 } { 2 } } d x } \\ { \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \operatorname { cosec } \theta ) ^ { - \frac { 1 } { 2 } } ( \cos - \theta ^ { \frac { 4 } { 3 } } ) d \theta } \end{array} \right. \)

Ask by Wood Barnett. in Kenya
Feb 20,2025

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Tutor-Verified Answer

Answer

These integrals can be evaluated using the Beta and Gamma functions.

Solution

It seems that the integrals provided cannot be calculated directly using the tools available. However, I can guide you through the process of solving each integral step by step. ### 1. Integral \( \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \sin x ) ^ { \frac { 8 } { 3 } } ( \sec x ) ^ { \frac { 1 } { 2 } } d x \) **Step 1:** Rewrite the integral using the identity \( \sec x = \frac{1}{\cos x} \): \[ \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \sin x ) ^ { \frac { 8 } { 3 } } ( \sec x ) ^ { \frac { 1 } { 2 } } d x = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \sin x ) ^ { \frac { 8 } { 3 } } ( \frac{1}{\cos x})^{\frac{1}{2}} d x \] **Step 2:** Simplify the integral: \[ = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \sin x ) ^ { \frac { 8 } { 3 } } ( \cos x )^{- \frac{1}{2}} d x \] **Step 3:** Use the substitution \( u = \sin x \), \( du = \cos x \, dx \): \[ = \int _ { 0 } ^ { 1 } u^{\frac{8}{3}} (1-u^2)^{- \frac{1}{2}} du \] This integral can be evaluated using the Beta function or by recognizing it as a form of the Gamma function. ### 2. Integral \( \int _ { 0 } ^ { \pi } ( \sin x ) ^ { \frac { 7 } { 2 } } \cos ^ { \frac { 3 } { 2 } } d x \) **Step 1:** Use the identity \( \cos^{\frac{3}{2}} x = (\cos x)^{\frac{3}{2}} \): \[ \int _ { 0 } ^ { \pi } ( \sin x ) ^ { \frac { 7 } { 2 } } ( \cos x )^{\frac{3}{2}} d x \] **Step 2:** This integral can be evaluated using the Beta function: \[ = 2 \int _ { 0 } ^ { \frac{\pi}{2}} ( \sin x )^{\frac{7}{2}} ( \cos x )^{\frac{3}{2}} d x \] ### 3. Integral \( \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \operatorname { cosec } \theta ) ^ { - \frac { 1 } { 2 } } ( \cos - \theta ^ { \frac { 4 } { 3 } } ) d \theta \) **Step 1:** Rewrite using \( \csc \theta = \frac{1}{\sin \theta} \): \[ \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \sin \theta )^{\frac{1}{2}} ( \cos(-\theta))^{\frac{4}{3}} d \theta \] **Step 2:** Since \( \cos(-\theta) = \cos(\theta) \): \[ = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \sin \theta )^{\frac{1}{2}} ( \cos \theta)^{\frac{4}{3}} d \theta \] This integral can also be evaluated using the Beta function. ### Conclusion To evaluate these integrals, you can use the properties of the Beta and Gamma functions. If you need further assistance with specific calculations or numerical evaluations, please let me know!

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Bonus Knowledge

Did you know that definite integrals, like those you have there, have been studied since the time of ancient Greeks? Mathematicians like Archimedes used them to find areas and volumes of shapes well before the advent of calculus as we know it today. The development of integral calculus in the 17th century by Isaac Newton and Gottfried Wilhelm Leibniz really took these concepts to a new level, allowing us to calculate and uncover relationships between different mathematical functions. Now, when tackling integrals, especially those with trigonometric functions like sine and cosine, a common mistake can be neglecting the range of the function’s values or not paying attention to the symmetry of the interval. A good tip is to try using substitutions to simplify the integral or convert it to a known standard form. Also, remember to keep an eye out for identities that might streamline your work—trigonometric identities can be especially handy here! Happy integrating!

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