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int(-pi//4)^(pi//4)sin^(2)x dx=...

`int_(-pi//4)^(pi//4)sin^(2)x dx=`

A

`(pi-4)/(8)`

B

`(pi-2)/(4)`

C

`(pi-2)/(2)`

D

`pi`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx \), we can follow these steps: ### Step 1: Use the identity for \(\sin^2 x\) We know that: \[ \sin^2 x = \frac{1 - \cos(2x)}{2} \] Thus, we can rewrite the integral as: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1 - \cos(2x)}{2} \, dx \] ### Step 2: Factor out the constant Since \(\frac{1}{2}\) is a constant, we can factor it out of the integral: \[ = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (1 - \cos(2x)) \, dx \] ### Step 3: Split the integral Now, we can split the integral into two parts: \[ = \frac{1}{2} \left( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx - \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2x) \, dx \right) \] ### Step 4: Evaluate the first integral The first integral, \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx\), is simply the length of the interval: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx = \left[ x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \] ### Step 5: Evaluate the second integral Now, we evaluate the second integral: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2x) \, dx \] The integral of \(\cos(2x)\) is \(\frac{1}{2} \sin(2x)\): \[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \] Thus, we have: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{2} \left( \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right) = \frac{1}{2} (1 - (-1)) = \frac{1}{2} \cdot 2 = 1 \] ### Step 6: Combine the results Now, we can combine the results: \[ = \frac{1}{2} \left( \frac{\pi}{2} - 1 \right) = \frac{\pi}{4} - \frac{1}{2} \] ### Final Answer Thus, the value of the integral is: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = \frac{\pi}{4} - \frac{1}{2} \]
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