定积分三角代换求解疑问：∫₀ᵃ dx/(a²+x²)^(3/2)（a>0）

Hey there! Let's walk through where your substitution went wrong and fix it step by step—trig substitutions can be tricky with those tiny details, so no worries about missing them at first.

First: Fixing the Denominator Expansion #

You picked the right substitution with $x = a\tan\theta$ for integrals involving $a^2 + x^2$, but made a small error when plugging into the denominator:

Your version: $\left(a^2 + a\tan\theta\right)^{\frac{3}{2}}$
Correct version: $\left(a^2 + (a\tan\theta)^2\right){\frac{3}{2}}$

Remember, $x = a\tan\theta$, so $x^2 = a^2\tan2\theta$. We can factor out $a^2$ from the denominator to simplify:
$$a^2 + x^2 = a^2\left(1 + \tan^2\theta\right)$$
And using the Pythagorean identity $1 + \tan^2\theta = \sec^2\theta$, this will make the rest of the calculation much smoother.

Second: Updating the Integration Limits #

Another critical piece you missed is adjusting the bounds when switching variables. Since we're moving from integrating with respect to $x$ to $\theta$, we need to map the original $x$-values to their corresponding $\theta$-values:

• When $x = 0$: $0 = a\tan\theta \implies \tan\theta = 0 \implies \theta = 0$
• When $x = a$: $a = a\tan\theta \implies \tan\theta = 1 \implies \theta = \frac{\pi}{4}$

This is why the correct integral uses bounds from $0$ to $\frac{\pi}{4}$ instead of $0$ to $a$—we're no longer working with the variable $x$!

Putting It All Together #

Let's rewrite the integral correctly now:
$$\int _0^{a\frac{dx}{\left(a}2+x^2\right){\frac{3}{2}}} = \int _0^{{\frac{\pi}{4}}\frac{a\sec}2\theta}{\left(a^{2\left(1+\tan}2\theta\right)\right)^{\frac{3}{2}}}d\theta$$

Simplify the denominator using the identity we noted earlier:
$$\left(a^2\sec2\theta\right)^{\frac{3}{2}} = a^3\sec3\theta$$
(We can safely factor out $a^3$ because $a > 0$, and $\sec\theta$ is positive in the interval $[0, \frac{\pi}{4}]$.)

Cancel terms in the numerator and denominator:
$$\int _0^{{\frac{\pi}{4}}\frac{a\sec}2\theta}{a^3\sec3\theta}d\theta = \frac{1}{a^2}\int _0^{\frac{\pi}{4}}\cos\theta d\theta$$

Finally, compute the straightforward cosine integral:
$$\frac{1}{a^{2}\left[\sin\theta\right]_0}{\frac{\pi}{4}} = \frac{1}{a^2}\left(\sin\frac{\pi}{4} - \sin0\right) = \frac{\sqrt{2}}{2a^2}$$

The two main missteps were forgetting to square $a\tan\theta$ in the denominator and not updating the integration bounds when switching variables—easy fixes once you spot them!

内容的提问来源于stack exchange，提问作者sktsasus