Problem 10
Question
Evaluate the integral. \(\int_{0}^{\frac{\pi}{2}} \cos ^{5} x \sqrt{\sin x} d x\)
Step-by-Step Solution
Verified Answer
The value of the given integral is \(\frac{1}{9}\)
1Step 1: Converting the cosine to sine
We can use the Pythagorean identity \(\cos^2x = 1 - \sin^2x\) to express \(\cos^5x\) in terms of \(\sin x\). We keep one \(\cos^2x\) term and convert the other terms, getting the expression \(cos^3x * cos^2x\). This yields \((1 - \sin^2x)^\frac{3}{2} * \sqrt{1 - \sin^2x}\).
2Step 2: Substituting the variable
Now perform a substitution to simplify the integral. Let \(u = \sin x\), so that \(du = \cos x dx\). The integral's limits change to 0 and 1 when u is substituted for \(\sin x\). The integral is now \(\int_{0}^{1} u^3 * (1 - u^2)^\frac{3}{2} du\).
3Step 3: Solving the integral
The given integral is a standard form of polynomial, which can be solved using power rule. Evaluate the integral by computing \( -\frac{1}{9}(1 - u^2)^\frac{9}{2}\) from 0 to 1.
4Step 4: Substituting the variable back
Now substitute back \(u = \sin x\) to get the result in terms of x. The evaluation of the integral from 0 to 1 gives \(-\frac{1}{9}(1 - \sin^2 x)^\frac{9}{2}\) evaluated from 0 to \(\frac{\pi}{2}\).
5Step 5: Evaluate the integral
Finally, evaluate the expression at the limits of the integral, we get the result as \(\frac{1}{9}\).
Key Concepts
Trigonometric IntegralsSubstitution MethodDefinite Integrals
Trigonometric Integrals
Trigonometric integrals involve integrating functions that include trigonometric functions like sine or cosine. These integrals often appear in calculus problems and can be seen in expressions such as \(\int \cos^{5}x \sqrt{\sin x}\,dx\). In the provided problem, a trigonometric function \(\cos^{5}x\) is integrated. This involves understanding how to manipulate and integrate powers of sine and cosine.
Many trigonometric integrals require using identities or substitutions to simplify the expressions. For example, in the exercise, the Pythagorean identity \(\cos^{2}x = 1 - \sin^{2}x\) was employed to rewrite the integral in terms of sine. This transformation makes it easier to handle the integration. Mastering these techniques is crucial for solving trigonometric integrals effectively.
Students should practice rewriting trigonometric expressions using identities and understanding when to apply them. Once the functions are expressed in simpler forms, the process of integration becomes much more manageable.
Many trigonometric integrals require using identities or substitutions to simplify the expressions. For example, in the exercise, the Pythagorean identity \(\cos^{2}x = 1 - \sin^{2}x\) was employed to rewrite the integral in terms of sine. This transformation makes it easier to handle the integration. Mastering these techniques is crucial for solving trigonometric integrals effectively.
Students should practice rewriting trigonometric expressions using identities and understanding when to apply them. Once the functions are expressed in simpler forms, the process of integration becomes much more manageable.
Substitution Method
The substitution method is a technique used in calculus to make integration simpler by changing variables. This method is particularly useful in cases where the integrand (the function being integrated) is complicated. In this problem, substitution helps transform the integral into a more familiar form.
The key to substitution is to identify a part of the integrand that can be replaced with a new variable. For example, by letting \(u = \sin x\), we change the integral's terms, making it easier to evaluate. The original integral then becomes \(\int_{0}^{1} u^{3}(1 - u^{2})^{\frac{3}{2}}\,du\), which is simpler than the original expression with respect to \(x\).
Once the substitution is made, it's also important to adjust the limits of integration. In our case, when \(x = 0\), \(u\) becomes 0, and when \(x = \frac{\pi}{2}\), \(u\) becomes 1. This is crucial as it avoids having to switch back to the original variable until after the integration process is complete.
Substitution is a powerful tool in solving integrals and forms the foundation for understanding more advanced integration techniques.
The key to substitution is to identify a part of the integrand that can be replaced with a new variable. For example, by letting \(u = \sin x\), we change the integral's terms, making it easier to evaluate. The original integral then becomes \(\int_{0}^{1} u^{3}(1 - u^{2})^{\frac{3}{2}}\,du\), which is simpler than the original expression with respect to \(x\).
Once the substitution is made, it's also important to adjust the limits of integration. In our case, when \(x = 0\), \(u\) becomes 0, and when \(x = \frac{\pi}{2}\), \(u\) becomes 1. This is crucial as it avoids having to switch back to the original variable until after the integration process is complete.
Substitution is a powerful tool in solving integrals and forms the foundation for understanding more advanced integration techniques.
Definite Integrals
Definite integrals are an essential concept in calculus, representing the accumulation of quantities over a specified interval. Evaluating definite integrals results in a number, in this case \(\frac{1}{9}\), as they measure the total change across the limits of integration.
In the problem, we evaluate the integral from \(0\) to \(\frac{\pi}{2}\). Once the function was simplified and the substitution was applied, it became a polynomial that was easily integrated using rules for powers.
To evaluate definite integrals, one substitutes the boundary values (upper and lower limits) into the antiderivative (the integrated function). The result is the net area or accumulation between these boundaries. This contrasts to indefinite integrals, which represent families of functions without specific limits.
Understanding how to properly evaluate definite integrals, especially involving trigonometric or algebraic substitutions, is critical. It helps calculate exact areas under curves, important in physics, engineering, and beyond. Definite integrals enable precise assessment of various scenarios requiring aggregation of distributive parts.
In the problem, we evaluate the integral from \(0\) to \(\frac{\pi}{2}\). Once the function was simplified and the substitution was applied, it became a polynomial that was easily integrated using rules for powers.
To evaluate definite integrals, one substitutes the boundary values (upper and lower limits) into the antiderivative (the integrated function). The result is the net area or accumulation between these boundaries. This contrasts to indefinite integrals, which represent families of functions without specific limits.
Understanding how to properly evaluate definite integrals, especially involving trigonometric or algebraic substitutions, is critical. It helps calculate exact areas under curves, important in physics, engineering, and beyond. Definite integrals enable precise assessment of various scenarios requiring aggregation of distributive parts.
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