© All Rights reserved @ LearnWithDash
Step-by-step Solution
Step 1: Identify the Integral
We are asked to evaluate
\int_{0}^{\sqrt{3}} \frac{15 \, x^3}{\sqrt{\,1 + x^2 + \sqrt{(1 + x^2)^3}\,}} \, dx
and then express the result in the form
\alpha \sqrt{2} + \beta \sqrt{3},
after which we must find the sum
\alpha + \beta.
Step 2: First Substitution ( x = \tan \theta )
Let
x = \tan \theta.
Then
dx = \sec^2 \theta \, d\theta,
and note that when
x = 0 \quad \rightarrow \quad \theta = 0,
and when
x = \sqrt{3} \quad \rightarrow \quad \theta = \frac{\pi}{3}.
So the integral becomes:
I \;=\; \int_{0}^{\frac{\pi}{3}} \frac{15 \,\tan^3 \theta \;\sec^2 \theta}{\sqrt{1 + \tan^2 \theta + \sqrt{(1 + \tan^2 \theta)^3}}} \, d\theta.
Step 3: Simplify Using 1 + \tan^2 \theta = \sec^2 \theta
Since 1 + \tan^2 \theta = \sec^2 \theta , we have inside the square root:
1 + \tan^2 \theta = \sec^2 \theta \quad \text{and} \quad \sqrt{(1 + \tan^2 \theta)^3} = \sqrt{\sec^6 \theta} = \sec^3 \theta.
Thus the integrand's denominator becomes
\sqrt{\sec^2 \theta + \sec^3 \theta} = \sqrt{\sec^2 \theta (1 + \sec \theta)} = \sec \theta \sqrt{1 + \sec \theta}.
We then rewrite the integral as:
I \;=\; \int_{0}^{\frac{\pi}{3}} \frac{15\,\tan^3 \theta \;\sec^2 \theta}{\sec \theta \,\sqrt{1 + \sec \theta}} \, d\theta
\;=\; \int_{0}^{\frac{\pi}{3}} \frac{15\,\tan^3 \theta \;\sec \theta}{\sqrt{1 + \sec \theta}} \, d\theta.
Step 4: Express \tan^3 \theta in a Useful Form
Note that
\tan^3 \theta = \tan \theta \;(\sec^2 \theta - 1)
because
\tan^2 \theta = \sec^2 \theta - 1.
Therefore,
\tan^3 \theta = \tan \theta \,(\sec^2 \theta - 1).
Substituting back gives:
I
= \int_{0}^{\frac{\pi}{3}} \frac{15 \,\tan \theta \,(\sec^2 \theta - 1)\,\sec \theta}{\sqrt{1 + \sec \theta}} \, d\theta.
Step 5: Second Substitution ( t^2 = 1 + \sec \theta )
To tackle the factor \sqrt{1 + \sec \theta} , set
t^2 = 1 + \sec \theta \quad \Longrightarrow \quad \sec \theta = t^2 - 1.
We also need dt in terms of \theta . Differentiating,
2t\, dt = \sec \theta \,\tan \theta \, d\theta \quad \Longrightarrow \quad \sec \theta \,\tan \theta \, d\theta = 2t \, dt.
Under this substitution, when
\theta = 0, \quad \sec 0 = 1 \quad \Longrightarrow \quad 1 + \sec 0 = 2 \quad \Longrightarrow \quad t = \sqrt{2},
and when
\theta = \frac{\pi}{3}, \quad \sec \left(\frac{\pi}{3}\right) = 2 \quad \Longrightarrow \quad 1 + 2 = 3 \quad \Longrightarrow \quad t = \sqrt{3}.
Hence, the limits of integration change from \theta \in \bigl[0, \frac{\pi}{3}\bigr] to t \in \bigl[\sqrt{2}, \sqrt{3}\bigr].
Step 6: Rewrite the Integral in Terms of t
Inside the integrand,
\sec^2 \theta - 1 = (\sec \theta)^2 - 1 = \sec^2 \theta - 1,
but since \sec \theta = t^2 - 1, we get:
\sec^2 \theta = (t^2 - 1)^2.
Also recall:
\sec \theta \,\tan \theta \, d\theta = 2t \, dt.
So,
I \;=\; \int_{\sqrt{2}}^{\sqrt{3}} \frac{15\, (\sec^2 \theta - 1)\,\sec \theta \,\tan \theta}{t} \, dt
\;=\; \int_{\sqrt{2}}^{\sqrt{3}} \frac{15\,( (t^2 - 1)^2 - 1 ) \, 2t}{t} \, dt,
since \sqrt{1 + \sec \theta} = \sqrt{t^2} = t (because t>0 on this interval). Simplify the factor \frac{2t}{t} = 2, obtaining:
I \;=\; 30 \int_{\sqrt{2}}^{\sqrt{3}} \bigl((t^2 - 1)^2 - 1\bigr)\, dt.
Step 7: Simplify the Expression Inside the Integral
First expand (t^2 - 1)^2 :
(t^2 - 1)^2 = t^4 - 2t^2 + 1.
Hence,
(t^2 - 1)^2 - 1 = t^4 - 2t^2 + 1 - 1 = t^4 - 2t^2.
So the integral simplifies to:
I
= 30 \int_{\sqrt{2}}^{\sqrt{3}} \bigl(t^4 - 2t^2\bigr)\, dt.
Step 8: Integrate Term-by-Term
We find the antiderivative:
\int \bigl(t^4 - 2t^2\bigr)\, dt
= \frac{t^5}{5} - \frac{2t^3}{3}.
Therefore,
I
= 30 \,\Biggl[\frac{t^5}{5} - \frac{2t^3}{3}\Biggr]_{\sqrt{2}}^{\sqrt{3}}
= 30 \left(\left[\frac{(\sqrt{3})^5}{5} - \frac{2(\sqrt{3})^3}{3}\right] \;-\; \left[\frac{(\sqrt{2})^5}{5} - \frac{2(\sqrt{2})^3}{3}\right]\right).
Step 9: Evaluate at the Limits
Compute powers carefully:
(\sqrt{3})^3 = 3\sqrt{3} and (\sqrt{3})^5 = 3\sqrt{3}\,\sqrt{3} = 3 \times 3 = 9\sqrt{3}.
(\sqrt{2})^3 = 2\sqrt{2} and (\sqrt{2})^5 = 4\sqrt{2}.
So,
\frac{(\sqrt{3})^5}{5} = \frac{9\sqrt{3}}{5},
\quad
\frac{2(\sqrt{3})^3}{3} = \frac{2 \cdot 3\sqrt{3}}{3} = 2\sqrt{3},
\frac{(\sqrt{2})^5}{5} = \frac{4\sqrt{2}}{5},
\quad
\frac{2(\sqrt{2})^3}{3} = \frac{2 \cdot 2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}.
Hence,
I
= 30 \Biggl[ \Bigl(\frac{9\sqrt{3}}{5} - 2\sqrt{3}\Bigr) - \Bigl(\frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3}\Bigr) \Biggr].
Step 10: Combine Like Terms
First simplify inside each parenthesis:
Inside the square-root-of-3 bracket:
\frac{9\sqrt{3}}{5} - 2\sqrt{3}
= \sqrt{3}\Bigl(\frac{9}{5} - 2\Bigr)
= \sqrt{3}\Bigl(\frac{9}{5} - \frac{10}{5}\Bigr)
= -\frac{1}{5}\sqrt{3}.
(But carefully check the final numeric factor once multiplied by 30.)
Inside the square-root-of-2 bracket:
\frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3}
= \sqrt{2}\left(\frac{4}{5} - \frac{4}{3}\right)
= 4\sqrt{2}\left(\frac{1}{5} - \frac{1}{3}\right)
= 4\sqrt{2}\left(\frac{3}{15} - \frac{5}{15}\right)
= 4\sqrt{2}\times\bigl(-\frac{2}{15}\bigr)
= -\frac{8\sqrt{2}}{15}.
Putting them together:
I
= 30\,\Bigl[\bigl(-\tfrac{1}{5}\sqrt{3}\bigr) - \bigl(-\tfrac{8\sqrt{2}}{15}\bigr)\Bigr]
= 30 \Bigl[ -\tfrac{1}{5}\sqrt{3} + \tfrac{8\sqrt{2}}{15} \Bigr].
Distribute 30:
I
= 30 \times \bigl(-\tfrac{1}{5}\sqrt{3}\bigr) \;+\; 30 \times \bigl(\tfrac{8\sqrt{2}}{15}\bigr)
= -6\sqrt{3} \;+\; 16\sqrt{2}.
Step 11: Identify \alpha and \beta and Compute their Sum
We have
I = 16\sqrt{2} - 6\sqrt{3},
so
\alpha = 16 \quad \text{and} \quad \beta = -6.
Therefore,
\alpha + \beta = 16 + \bigl(-6\bigr) = 10.
Final Answer
The value of
\alpha + \beta
is
\boxed{10}.
