To solve the equation
\[
\int_0^{\frac{\pi}{4}} \sec^2 x \sin x \, dx = \int_0^a \frac{dx}{\sqrt{x} + \sqrt{x + a}},
\]
we will evaluate both sides step by step.
### Step 1: Evaluate the Left-Hand Side (LHS)
The LHS is given by:
\[
\int_0^{\frac{\pi}{4}} \sec^2 x \sin x \, dx.
\]
We can rewrite \(\sec^2 x\) as \(\frac{1}{\cos^2 x}\) and \(\sin x\) remains as it is:
\[
\int_0^{\frac{\pi}{4}} \sec^2 x \sin x \, dx = \int_0^{\frac{\pi}{4}} \frac{\sin x}{\cos^2 x} \, dx.
\]
Using the identity \(\tan x = \frac{\sin x}{\cos x}\), we can express this integral as:
\[
\int_0^{\frac{\pi}{4}} \tan x \sec x \, dx.
\]
### Step 2: Integrate the LHS
The integral of \(\tan x \sec x\) is \(\sec x\):
\[
\int \tan x \sec x \, dx = \sec x + C.
\]
Now we evaluate it from \(0\) to \(\frac{\pi}{4}\):
\[
\left[ \sec x \right]_0^{\frac{\pi}{4}} = \sec\left(\frac{\pi}{4}\right) - \sec(0).
\]
Calculating the values:
\[
\sec\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \sec(0) = 1.
\]
Thus, the LHS becomes:
\[
\sqrt{2} - 1.
\]
### Step 3: Evaluate the Right-Hand Side (RHS)
The RHS is given by:
\[
\int_0^a \frac{dx}{\sqrt{x} + \sqrt{x + a}}.
\]
To simplify this integral, we rationalize the denominator:
\[
\frac{1}{\sqrt{x} + \sqrt{x + a}} \cdot \frac{\sqrt{x + a} - \sqrt{x}}{\sqrt{x + a} - \sqrt{x}} = \frac{\sqrt{x + a} - \sqrt{x}}{(x + a) - x} = \frac{\sqrt{x + a} - \sqrt{x}}{a}.
\]
Thus, we have:
\[
\int_0^a \frac{dx}{\sqrt{x} + \sqrt{x + a}} = \frac{1}{a} \int_0^a (\sqrt{x + a} - \sqrt{x}) \, dx.
\]
### Step 4: Evaluate the Integral
Now we evaluate:
\[
\int_0^a \sqrt{x + a} \, dx - \int_0^a \sqrt{x} \, dx.
\]
1. For \(\int_0^a \sqrt{x + a} \, dx\):
Using the substitution \(u = x + a\), \(du = dx\):
\[
\int_0^a \sqrt{x + a} \, dx = \int_a^{2a} \sqrt{u} \, du = \left[\frac{2}{3} u^{3/2}\right]_a^{2a} = \frac{2}{3} \left((2a)^{3/2} - a^{3/2}\right).
\]
2. For \(\int_0^a \sqrt{x} \, dx\):
\[
\int_0^a \sqrt{x} \, dx = \left[\frac{2}{3} x^{3/2}\right]_0^a = \frac{2}{3} a^{3/2}.
\]
### Step 5: Combine the Results
Combining both integrals:
\[
\int_0^a \sqrt{x + a} \, dx - \int_0^a \sqrt{x} \, dx = \frac{2}{3} \left( (2a)^{3/2} - a^{3/2} \right) - \frac{2}{3} a^{3/2} = \frac{2}{3} \left( 2^{3/2} a^{3/2} - 2a^{3/2} \right) = \frac{2}{3} (2\sqrt{2} - 2) a^{3/2}.
\]
Thus, we have:
\[
\int_0^a \frac{dx}{\sqrt{x} + \sqrt{x + a}} = \frac{1}{a} \cdot \frac{2}{3} (2\sqrt{2} - 2) a^{3/2} = \frac{2}{3} (2\sqrt{2} - 2) \sqrt{a}.
\]
### Step 6: Equate LHS and RHS
Now we equate the LHS and RHS:
\[
\sqrt{2} - 1 = \frac{2}{3} (2\sqrt{2} - 2) \sqrt{a}.
\]
### Step 7: Solve for \(a\)
To find \(a\), we rearrange:
\[
\sqrt{a} = \frac{3(\sqrt{2} - 1)}{2(2\sqrt{2} - 2)} = \frac{3(\sqrt{2} - 1)}{4(\sqrt{2} - 1)} = \frac{3}{4}.
\]
Squaring both sides gives:
\[
a = \left(\frac{3}{4}\right)^2 = \frac{9}{16}.
\]
### Final Answer
Thus, the value of \(a\) is:
\[
\alpha = \frac{9}{16}.
\]
