Q3:

Evaluate the integrals in Exercises 1 to 8 using substitution. 3. $\int_{0}^{1} \sin^{-1} (\frac{2x}{1+x^2}) dx$

Step 1: Simplifying the Integrand using Trigonometric Substitution

The integrand contains the inverse trigonometric expression $\sin^{-1} \left( \frac{2x}{1+x^2} \right)$. We utilize the substitution $x = \tan \theta$. Consequently, $dx = \sec^2 \theta \, d\theta$. [By the substitution rule for integrals].

Under this transformation, the expression simplifies using the double-angle identity for sine: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$. Thus, $\sin^{-1} \left( \frac{2\tan\theta}{1+\tan^2\theta} \right) = \sin^{-1}(\sin 2\theta) = 2\theta$, provided $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Given $x \in [0, 1]$, $\theta \in [0, \frac{\pi}{4}]$, satisfying the domain constraint.

Step 2: Adjusting Limits of Integration

We must transform the limits from $x$-space to $\theta$-space:

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

Step 3: Setting up the Transformed Integral

Substituting the variables into the original integral $I = \int_{0}^{1} \sin^{-1} \left( \frac{2x}{1+x^2} \right) dx$:

$I = \int_{0}^{\pi/4} (2\theta) \sec^2 \theta \, d\theta = 2 \int_{0}^{\pi/4} \theta \sec^2 \theta \, d\theta$

Step 4: Integration by Parts

Applying the formula $\int u \, dv = uv - \int v \, du$, let $u = \theta$ and $dv = \sec^2 \theta \, d\theta$. Then $du = d\theta$ and $v = \tan \theta$. [Per Integration by Parts theorem].

$I = 2 \left[ \theta \tan \theta - \int \tan \theta \, d\theta \right]_{0}^{\pi/4}$

$I = 2 \left[ \theta \tan \theta - \ln|\sec \theta| \right]_{0}^{\pi/4}$

Step 5: Evaluating at the Limits

Applying the Fundamental Theorem of Calculus:

$I = 2 \left[ \left( \frac{\pi}{4} \tan \frac{\pi}{4} - \ln|\sec \frac{\pi}{4}| \right) - (0 \cdot \tan 0 - \ln|\sec 0|) \right]$

Since $\tan \frac{\pi}{4} = 1$, $\sec \frac{\pi}{4} = \sqrt{2}$, $\tan 0 = 0$, and $\sec 0 = 1$ (noting $\ln(1) = 0$):

$I = 2 \left[ \frac{\pi}{4}(1) - \ln(\sqrt{2}) - 0 \right]$

$I = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \ln(2) \right]$

$I = \frac{\pi}{2} - \ln(2)$

Final Solution: $\frac{\pi}{2} - \ln(2)$