If `int_(0)^(50pi) (sin^(4) x +cos^(4) x)dx = k int_(0)^(pi//2) ((3)/(4) + (1)/(4) cos 4x)dx`, then k=

Options

A. 200
B. 100
C. 50
D. 25

Answer

To solve the given problem, we start with the equation:

\[ \int_{0}^{50\pi} (\sin^4 x + \cos^4 x) \, dx = k \int_{0}^{\frac{\pi}{2}} \left( \frac{3}{4} + \frac{1}{4} \cos 4x \right) \, dx \]

### Step 1: Evaluate the left-hand side integral

The function \(\sin^4 x + \cos^4 x\) can be simplified using the identity:

\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]

Using the identity \(\sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x\), we have:

\[ \sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2 2x \]

Thus, we can rewrite the integral:

\[ \int_{0}^{50\pi} (\sin^4 x + \cos^4 x) \, dx = \int_{0}^{50\pi} \left( 1 - \frac{1}{2} \sin^2 2x \right) \, dx \]

### Step 2: Split the integral

Now, we can split the integral into two parts:

\[ \int_{0}^{50\pi} 1 \, dx - \frac{1}{2} \int_{0}^{50\pi} \sin^2 2x \, dx \]

Calculating the first integral:

\[ \int_{0}^{50\pi} 1 \, dx = 50\pi \]

### Step 3: Evaluate the second integral

To evaluate \(\int_{0}^{50\pi} \sin^2 2x \, dx\), we use the property of periodic functions. The period of \(\sin^2 2x\) is \(\frac{\pi}{2}\). Therefore, we can calculate:

\[ \int_{0}^{50\pi} \sin^2 2x \, dx = 100 \int_{0}^{\frac{\pi}{2}} \sin^2 2x \, dx \]

Using the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\):

\[ \int_{0}^{\frac{\pi}{2}} \sin^2 2x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 4x}{2} \, dx = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} \]

Thus:

\[ \int_{0}^{50\pi} \sin^2 2x \, dx = 100 \cdot \frac{\pi}{4} = 25\pi \]

### Step 4: Combine results

Now substituting back, we have:

\[ \int_{0}^{50\pi} (\sin^4 x + \cos^4 x) \, dx = 50\pi - \frac{1}{2} \cdot 25\pi = 50\pi - 12.5\pi = 37.5\pi \]

### Step 5: Evaluate the right-hand side integral

Now we evaluate the right-hand side:

\[ \int_{0}^{\frac{\pi}{2}} \left( \frac{3}{4} + \frac{1}{4} \cos 4x \right) \, dx = \int_{0}^{\frac{\pi}{2}} \frac{3}{4} \, dx + \int_{0}^{\frac{\pi}{2}} \frac{1}{4} \cos 4x \, dx \]

Calculating the first integral:

\[ \int_{0}^{\frac{\pi}{2}} \frac{3}{4} \, dx = \frac{3}{4} \cdot \frac{\pi}{2} = \frac{3\pi}{8} \]

For the second integral:

\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{4} \cos 4x \, dx = \frac{1}{4} \cdot \left[ \frac{\sin 4x}{4} \right]_{0}^{\frac{\pi}{2}} = \frac{1}{4} \cdot \left(0 - 0\right) = 0 \]

Thus, the right-hand side becomes:

\[ \int_{0}^{\frac{\pi}{2}} \left( \frac{3}{4} + \frac{1}{4} \cos 4x \right) \, dx = \frac{3\pi}{8} \]

### Step 6: Set up the equation for k

Now we have:

\[ 37.5\pi = k \cdot \frac{3\pi}{8} \]

Dividing both sides by \(\pi\):

\[ 37.5 = k \cdot \frac{3}{8} \]

Multiplying both sides by \(\frac{8}{3}\):

\[ k = 37.5 \cdot \frac{8}{3} = 100 \]

Thus, the value of \(k\) is:

\[ \boxed{100} \]