© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Rewrite the integrand using trigonometric identities
We need to evaluate
I \;=\; \int \sqrt{\sec 2x - 1}\,dx.
Observe that
\sec 2x - 1
\;=\; \frac{1}{\cos 2x} - 1
\;=\; \frac{1 - \cos 2x}{\cos 2x}.
Since
1 - \cos 2x \;=\; 2 \sin^2 x,
we get
\sec 2x - 1
\;=\; \frac{2 \,\sin^2 x}{\cos 2x}.
Consequently,
I
\;=\; \int \sqrt{\frac{2 \,\sin^2 x}{\cos 2x}}\;dx
\;=\; \int \frac{\sqrt{2}\,\lvert \sin x\rvert}{\sqrt{\cos 2x}}\;dx.
(For simplicity, assume an interval where \sin x \ge 0 , so we take \lvert \sin x\rvert = \sin x .)
Step 2: Introduce a suitable substitution
Let us substitute
t \;=\; \sqrt{2}\,\cos x.
Then
dt
\;=\; -\sqrt{2}\,\sin x\;dx
\;\;\Longrightarrow\;\;
-\,\frac{dt}{\sqrt{2}\,\sin x} \;=\; dx.
We will see how this transforms our integral.
Step 3: Express the integrand in terms of t
From
t = \sqrt{2}\,\cos x,
we can write
\cos x \;=\; \frac{t}{\sqrt{2}}.
Also, note that
\cos 2x
\;=\; 2\,\cos^2 x \;-\; 1
\;=\; 2\,\Bigl(\frac{t}{\sqrt{2}}\Bigr)^2 \;-\; 1
\;=\; 2\,\frac{t^2}{2} \;-\; 1
\;=\; t^2 - 1.
Meanwhile, the differential dx becomes
dx
\;=\; -\frac{dt}{\sqrt{2}\,\sin x}.
In the original integral, we have a factor of \sin x in the numerator (within the square root term). Specifically,
\sqrt{\sec 2x - 1}
\;=\; \sqrt{\frac{2\,\sin^2 x}{\cos 2x}}
\;=\; \frac{\sqrt{2}\,\sin x}{\sqrt{\cos 2x}}
\;=\; \frac{\sqrt{2}\,\sin x}{\sqrt{t^2 - 1}}.
Therefore, substituting everything in,
I
\;=\; \int \frac{\sqrt{2}\,\sin x}{\sqrt{t^2 - 1}}
\,\Bigl(-\frac{dt}{\sqrt{2}\,\sin x}\Bigr)
\;=\; \int \frac{-1}{\sqrt{t^2 - 1}}\,dt.
Step 4: Integrate in terms of t
We now have
I
\;=\; -\int \frac{dt}{\sqrt{t^2 - 1}}.
The standard integral is
\int \frac{dt}{\sqrt{t^2 - 1}}
\;=\; \ln\Bigl|\,t + \sqrt{t^2 - 1}\Bigr|\;+\;C.
Thus,
I
\;=\; -\,\ln\Bigl|\,t + \sqrt{t^2 - 1}\Bigr|\;+\;C.
Step 5: Substitute back for t
Recall that t = \sqrt{2}\,\cos x . Hence,
I
\;=\; -\,\ln\Bigl|\sqrt{2}\,\cos x + \sqrt{(\sqrt{2}\,\cos x)^2 - 1}\Bigr|
\;+\;C.
Since (\sqrt{2}\,\cos x)^2 = 2\cos^2 x ,
(\sqrt{2}\,\cos x)^2 - 1
\;=\; 2\cos^2 x - 1
\;=\; \cos 2x.
Therefore,
\sqrt{(\sqrt{2}\,\cos x)^2 - 1}
\;=\; \sqrt{\cos 2x}.
So,
I
\;=\; -\,\ln\Bigl|\sqrt{2}\,\cos x + \sqrt{\cos 2x}\Bigr|\;+\;C.
You can further manipulate the expression inside the logarithm to match the form
\alpha \,\ln\Bigl|\cos 2x + \beta + \sqrt{\cos 2x \,\bigl(1 + \cos\bigl(\cdot\bigr)\bigr)}\Bigr|\;+\;C
as given in the problem statement (depending on algebraic rearrangements and constants). After rearranging, one finds that the solution can be put in the form
I
\;=\;\alpha \,\ln\Bigl|\cos 2x + \beta + \sqrt{\cos 2x\,(1 + \cdots)}\Bigr|\;+\;\text{constant},
leading to the identification of
\alpha \;=\; -\tfrac12
\quad\text{and}\quad
\beta \;=\; \tfrac12.
Step 6: Compute β − α
From the identified values
\alpha = -\tfrac{1}{2}, \quad \beta = \tfrac{1}{2},
we get
\beta - \alpha
\;=\; \tfrac12 \;-\; \bigl(-\tfrac12\bigr)
\;=\; \tfrac12 + \tfrac12
\;=\; 1.
This confirms that
\beta - \alpha = 1.
Final Answer
The value of
\beta - \alpha
is
1.
