Substitution — Second Type (Trigonometric)

Approach: The denominator is a positive integer power of $1+x^2$, so put $x = \tan t$ and use the identity $1 + \tan^2 t = \sec^2 t$ to turn the denominator into $\sec^3 t$; the $\sec^2 t$ in $dx$ then cancels one factor of $\sec t$. The validity condition is that $t$ must be taken in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, where $\sec t > 0$, so that $(1+x^2)^{3/2} = \sec^3 t$ holds (otherwise one would need $|\sec t|$ cubed).

Steps:

1. Let $x = \tan t$ with $t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so $dx = \sec^2 t \, dt$ and $1 + x^2 = 1 + \tan^2 t = \sec^2 t$, hence $(1+x^2)^{3/2} = \sec^3 t$.
2. Substitute: $\int \frac{dx}{(1+x^2)^{3/2}} = \int \frac{\sec^2 t}{\sec^3 t} \, dt = \int \cos t \, dt = \sin t + C$.
3. Back-substitute: from $\tan t = x$ we get $\sin t = \frac{x}{\sqrt{1+x^2}}$, so the original integral is $\frac{x}{\sqrt{1+x^2}} + C$.
4. Verify by differentiating: $\frac{d}{dx}\left[\frac{x}{\sqrt{1+x^2}}\right] = \frac{1}{\sqrt{1+x^2}} - \frac{x^2}{(1+x^2)^{3/2}} = \frac{1+x^2-x^2}{(1+x^2)^{3/2}} = \frac{1}{(1+x^2)^{3/2}}$, which matches the original integrand.

Answer: $\int \frac{dx}{(1+x^2)^{3/2}} = \frac{x}{\sqrt{1+x^2}} + C$.

Approach: When a substitution is made in a definite integral, the limits must be transformed along with the integrand. The new upper limit is fixed by $x = 1$, which gives $\tan t = 1$ and hence $t = \frac{\pi}{4}$; the new lower limit is fixed by $x = 0$, which gives $t = 0$. This step is the most common place to lose marks.

Steps:

1. Let $x = \tan t$ with $t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so $dx = \sec^2 t \, dt$ and $1+x^2 = 1+\tan^2 t = \sec^2 t$.
2. Transform the limits: $x = 0$ corresponds to $\tan t = 0$, that is $t = 0$; $x = 1$ corresponds to $\tan t = 1$, that is $t = \frac{\pi}{4}$.
3. Substitute and integrate: $\int_0^1 \frac{dx}{1+x^2} = \int_0^{\pi/4} \frac{\sec^2 t}{\sec^2 t} \, dt = \int_0^{\pi/4} dt = \left[t\right]_0^{\pi/4} = \frac{\pi}{4}$.
4. Cross-check with the Newton–Leibniz formula: the antiderivative is $\arctan x$, so $\int_0^1 \frac{dx}{1+x^2} = \arctan 1 - \arctan 0 = \frac{\pi}{4} - 0 = \frac{\pi}{4}$, in agreement with step 3.

Answer: $\int_0^1 \frac{dx}{1+x^2} = \frac{\pi}{4}$.

Approach: The radical has the form $\sqrt{x^2-a^2}$ (here $a = 1$), so put $x = a\sec t$ and use the identity $\sec^2 t - 1 = \tan^2 t$. The valid interval is determined by $x \ge a$: taking $t \in \left[0, \frac{\pi}{2}\right)$ gives $\sec t \ge 1$ and $\tan t \ge 0$, so $x = \sec t \ge 1$ and $\sqrt{x^2-1} = \tan t$ holds (otherwise one would need $|\tan t|$).

Steps:

1. Let $x = \sec t$ with $t \in \left[0, \frac{\pi}{2}\right)$, so $dx = \sec t \tan t \, dt$ and $\sqrt{x^2-1} = \sqrt{\sec^2 t - 1} = \sqrt{\tan^2 t} = \tan t$ (here $\tan t \ge 0$).
2. Substitute: $\int \frac{dx}{\sqrt{x^2-1}} = \int \frac{\sec t \tan t}{\tan t} \, dt = \int \sec t \, dt = \ln|\sec t + \tan t| + C$.
3. Back-substitute: from $\sec t = x$ and $\tan t = \sqrt{x^2-1}$ we get $\ln|\sec t + \tan t| = \ln|x + \sqrt{x^2-1}|$.
4. Verify by differentiating (for $x > 1$ we have $x + \sqrt{x^2-1} > 0$): $\frac{d}{dx}\ln\left(x + \sqrt{x^2-1}\right) = \frac{1 + \frac{x}{\sqrt{x^2-1}}}{x + \sqrt{x^2-1}} = \frac{\sqrt{x^2-1} + x}{\sqrt{x^2-1}\left(x + \sqrt{x^2-1}\right)} = \frac{1}{\sqrt{x^2-1}}$, which matches the original integrand.

Answer: $\int \frac{dx}{\sqrt{x^2-1}} = \ln\left|x + \sqrt{x^2-1}\right| + C$.