(b) Use the substitution $$ x=\sin \theta $$ to show that $$ \int \frac{\cos \theta}{\sqrt{2 \cos 2 \theta-1}} \mathrm{~d} \theta=\int \frac{1}{\sqrt{1-4 x^{2}}} \mathrm{~d} x $$ Given that $$ \int_{0}^{\alpha} \frac{\cos \theta}{\sqrt{2 \cos 2 \theta-1}} \mathrm{~d} \theta=\frac{\pi}{4} $$ find the value of $$ \alpha $$.

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We start by making the substitution
$$
x = \sin \theta.
$$
Then
$$
\mathrm{d}x = \cos \theta \, \mathrm{d}\theta,
$$
so that the integral
$$
\int \frac{\cos \theta}{\sqrt{2\cos2\theta - 1}} \, \mathrm{d}\theta
$$
becomes
$$
\int \frac{1}{\sqrt{2\cos2\theta - 1}} \, \mathrm{d}x.
$$

Next, recall the double-angle formula for cosine:
$$
\cos2\theta = 1 - 2\sin^2\theta.
$$
Substitute this into the expression under the square root:
$$
2\cos2\theta - 1 = 2\left(1 - 2\sin^2\theta\right) - 1 = 2 - 4\sin^2\theta - 1 = 1 - 4\sin^2\theta.
$$
Since $$x = \sin \theta$$, it becomes
$$
1 - 4\sin^2\theta = 1 - 4x^2.
$$
Thus, we have
$$
\int \frac{\cos \theta}{\sqrt{2\cos 2\theta - 1}} \, \mathrm{d}\theta = \int \frac{1}{\sqrt{1 - 4x^2}} \, \mathrm{d}x.
$$

Now, we are given the definite integral
$$
\int_{0}^{\alpha} \frac{\cos \theta}{\sqrt{2\cos 2\theta-1}} \mathrm{~d}\theta = \frac{\pi}{4}.
$$
Changing the variable with $$x = \sin \theta$$ changes the limits. When $$\theta = 0$$, $$x = \sin0 = 0$$, and when $$\theta = \alpha$$, $$x = \sin\alpha$$. Therefore, the integral becomes
$$
\int_{0}^{\sin \alpha} \frac{1}{\sqrt{1-4x^2}} \,\mathrm{d}x = \frac{\pi}{4}.
$$

To evaluate the integral on the left, let
$$
u = 2x \quad \text{so that} \quad \mathrm{d}x = \frac{1}{2}\mathrm{d}u.
$$
Then, the integral transforms into
$$
\int \frac{1}{\sqrt{1 - u^2}} \left(\frac{1}{2} \, \mathrm{d}u\right) = \frac{1}{2} \int \frac{\mathrm{d}u}{\sqrt{1-u^2}}.
$$
The antiderivative is well known:
$$
\int \frac{\mathrm{d}u}{\sqrt{1-u^2}} = \arcsin u + C.
$$
Thus, the indefinite integral becomes
$$
\frac{1}{2}\arcsin u + C = \frac{1}{2}\arcsin(2x) + C.
$$

Applying the limits, we have
$$
\frac{1}{2} \arcsin(2\sin\alpha) - \frac{1}{2}\arcsin(0) = \frac{1}{2}\arcsin(2\sin\alpha) = \frac{\pi}{4}.
$$
Multiplying both sides by 2, we find
$$
\arcsin(2\sin\alpha) = \frac{\pi}{2}.
$$
Taking the sine of both sides gives
$$
2\sin\alpha = \sin\left(\frac{\pi}{2}\right) = 1.
$$
Thus,
$$
\sin\alpha = \frac{1}{2}.
$$

Assuming $$\alpha$$ is in the first quadrant (which is typical in these problems), we have
$$
\alpha = \frac{\pi}{6}.
$$
$$ \alpha = \frac{\pi}{6} $$

(b) Use the substitution to show that

Given that

find the value of .

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(b) Use the substitution to show that Given that find the value of .