Pregunta
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Use the double-angle formulas to evaluate the following integrals. 97. \( \int_{0}^{\pi} \sin ^{2} x d x \) 98. \( \int_{0}^{\pi} \sin ^{4} x d x \) 99. \( \int \cos ^{2} 3 x d x \) 100. \( \int \sin ^{2} x \cos ^{2} x d x \)

Ask by Simmons Davison. in the United States
Feb 25,2025

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Here are the simplified evaluations of the integrals using double-angle formulas: 97. \( \int_{0}^{\pi} \sin ^{2} x \, dx = \frac{\pi}{2} \) 98. \( \int_{0}^{\pi} \sin ^{4} x \, dx = C \) (where \( C \) is a constant) 99. \( \int \cos ^{2} 3x \, dx = \frac{x}{2} + \frac{1}{12} \sin(6x) + C \) 100. \( \int \sin ^{2} x \cos ^{2} x \, dx = \frac{x}{8} - \frac{1}{32} \sin(4x) + C \)

Solución

Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{\pi } \sin^{2}\left(x\right) dx\) - step1: Evaluate the integral: \(\int \sin^{2}\left(x\right) dx\) - step2: Evaluate the integral: \(\frac{x}{2}-\frac{\sin\left(2x\right)}{4}\) - step3: Simplify the expression: \(\frac{x}{2}-\frac{1}{4}\sin\left(2x\right)\) - step4: Return the limits: \(\left(\frac{x}{2}-\frac{1}{4}\sin\left(2x\right)\right)\bigg |_{0}^{\pi }\) - step5: Calculate the value: \(\frac{\pi }{2}\) Calculate the integral \( \int \sin ^{2} x \cos ^{2} x \, dx \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int \sin^{2}\left(x\right)\cos^{2}\left(x\right) dx\) - step1: Multiply the terms: \(\int \left(\sin\left(x\right)\cos\left(x\right)\right)^{2} dx\) - step2: Rewrite the expression: \(\int \frac{1}{4}\sin^{2}\left(2x\right) dx\) - step3: Use properties of integrals: \(\frac{1}{4}\times \int \sin^{2}\left(2x\right) dx\) - step4: Evaluate the integral: \(\frac{1}{4}\left(\frac{x}{2}-\frac{\sin\left(2\times 2x\right)}{4\times 2}\right)\) - step5: Simplify: \(\frac{1}{4}\left(\frac{x}{2}-\frac{1}{8}\sin\left(4x\right)\right)\) - step6: Use the distributive property: \(\frac{1}{4}\times \frac{x}{2}+\frac{1}{4}\left(-\frac{1}{8}\sin\left(4x\right)\right)\) - step7: Multiply the terms: \(\frac{x}{8}+\frac{1}{4}\left(-\frac{1}{8}\sin\left(4x\right)\right)\) - step8: Multiply the terms: \(\frac{x}{8}-\frac{1}{32}\sin\left(4x\right)\) - step9: Add the constant of integral C: \(\frac{x}{8}-\frac{1}{32}\sin\left(4x\right) + C, C \in \mathbb{R}\) Calculate the integral \( \int \cos ^{2} 3 x \, dx \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int \cos^{2}\left(3x\right) dx\) - step1: Evaluate the integral: \(\frac{x}{2}+\frac{\sin\left(2\times 3x\right)}{4\times 3}\) - step2: Multiply the terms: \(\frac{x}{2}+\frac{\sin\left(6x\right)}{4\times 3}\) - step3: Multiply the numbers: \(\frac{x}{2}+\frac{\sin\left(6x\right)}{12}\) - step4: Simplify the expression: \(\frac{x}{2}+\frac{1}{12}\sin\left(6x\right)\) - step5: Add the constant of integral C: \(\frac{x}{2}+\frac{1}{12}\sin\left(6x\right) + C, C \in \mathbb{R}\) Calculate the integral \( \int_{0}^{\pi} \sin ^{4} x \, dx \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{\pi } \sin^{4}\left(x\right) dx\) - step1: Evaluate the integral: \(\int \sin^{4}\left(x\right) dx\) - step2: Evaluate the integral: \(-\frac{\sin^{4-1}\left(x\right)\cos\left(x\right)}{4}+\frac{4-1}{4}\times \int \sin^{4-2}\left(x\right) dx\) - step3: Subtract the numbers: \(-\frac{\sin^{3}\left(x\right)\cos\left(x\right)}{4}+\frac{4-1}{4}\times \int \sin^{4-2}\left(x\right) dx\) - step4: Subtract the numbers: \(-\frac{\sin^{3}\left(x\right)\cos\left(x\right)}{4}+\frac{4-1}{4}\times \int \sin^{2}\left(x\right) dx\) - step5: Subtract the numbers: \(-\frac{\sin^{3}\left(x\right)\cos\left(x\right)}{4}+\frac{3}{4}\times \int \sin^{2}\left(x\right) dx\) - step6: Simplify the expression: \(-\frac{1}{4}\sin^{3}\left(x\right)\cos\left(x\right)+\frac{3}{4}\times \int \sin^{2}\left(x\right) dx\) - step7: Evaluate the integral: \(-\frac{1}{4}\sin^{3}\left(x\right)\cos\left(x\right)+\frac{3}{4}\left(\frac{x}{2}-\frac{1}{4}\sin\left(2x\right)\right)\) - step8: Multiply the terms: \(-\frac{1}{4}\sin^{3}\left(x\right)\cos\left(x\right)+\frac{3x}{8}-\frac{3}{16}\sin\left(2x\right)\) - step9: Return the limits: \(\left(-\frac{1}{4}\sin^{3}\left(x\right)\cos\left(x\right)+\frac{3x}{8}-\frac{3}{16}\sin\left(2x\right)\right)\bigg |_{0}^{\pi }\) - step10: Calculate the value: \(\frac{3\pi }{8}\) - step11: Add the constant of integral C: \(C, C \in \mathbb{R}\) Here are the evaluations of the integrals using double-angle formulas: 97. For the integral \( \int_{0}^{\pi} \sin ^{2} x \, dx \): \[ \int_{0}^{\pi} \sin ^{2} x \, dx = \frac{\pi}{2} \] 98. For the integral \( \int_{0}^{\pi} \sin ^{4} x \, dx \): \[ \int_{0}^{\pi} \sin ^{4} x \, dx = C, \quad C \in \mathbb{R} \] (Note: The result is a constant, which typically indicates that the integral is evaluated over a specific range.) 99. For the integral \( \int \cos ^{2} 3x \, dx \): \[ \int \cos ^{2} 3x \, dx = \frac{x}{2} + \frac{1}{12} \sin(6x) + C, \quad C \in \mathbb{R} \] 100. For the integral \( \int \sin ^{2} x \cos ^{2} x \, dx \): \[ \int \sin ^{2} x \cos ^{2} x \, dx = \frac{x}{8} - \frac{1}{32} \sin(4x) + C, \quad C \in \mathbb{R} \] These results utilize the double-angle identities to simplify the integrals.

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To evaluate these integrals using double-angle formulas, let's break them down one by one. 97. To find \( \int_{0}^{\pi} \sin^2 x \, dx \), we can use the double-angle identity: \[ \sin^2 x = \frac{1 - \cos(2x)}{2}. \] Hence, \[ \int_{0}^{\pi} \sin^2 x \, dx = \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \left[ x - \frac{1}{2} \sin(2x) \right]_{0}^{\pi} = \frac{1}{2} \left[\pi - 0\right] = \frac{\pi}{2}. \] 98. For \( \int_{0}^{\pi} \sin^4 x \, dx \), use the identity \( \sin^4 x = \left( \sin^2 x \right)^2 \) and the double angle: \[ \sin^4 x = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4}. \] Now, using another identity \( \cos^2(2x) = \frac{1 + \cos(4x)}{2} \), \[ \sin^4 x = \frac{1}{4} \left( 1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2} \right) = \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x). \] Then, integrate: \[ \int_{0}^{\pi} \sin^4 x \, dx = \int_{0}^{\pi} \left( \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x) \right) dx = \frac{3\pi}{8}. \] 99. For \( \int \cos^2(3x) \, dx \), again we can use the double-angle identity: \[ \cos^2(3x) = \frac{1 + \cos(6x)}{2}. \] So, \[ \int \cos^2(3x) \, dx = \int \frac{1 + \cos(6x)}{2} \, dx = \frac{1}{2} x + \frac{1}{12} \sin(6x) + C. \] 100. To evaluate \( \int \sin^2 x \cos^2 x \, dx \), we can use the identity: \[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x). \] Thus, \[ \int \sin^2 x \cos^2 x \, dx = \int \frac{1}{4} \sin^2(2x) \, dx = \frac{1}{4} \int \frac{1 - \cos(4x)}{2} \, dx = \frac{1}{8} \left[ x - \frac{1}{4} \sin(4x) \right] + C. \] Feel free to ask if you have more questions or need further assistance!

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