Question

Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

Hint:

Always change your limits of integration immediately when performing a substitution in a definite integral so you don't have to substitute back to the original variable at the end.

Answer 1

Concept: Definite integrals involving inverse trigonometric functions are often simplified through trigonometric substitution. For expressions containing \( 1 + x^2 \), substituting \( x = \tan\theta \) is highly effective because it simplifies the denominator via the identity \( 1 + \tan^2\theta = \sec^2\theta \).

Step 1: Applying trigonometric substitution and altering the integration limits.
Let the given integral be: \[ I = \int_{0}^{1} \frac{x \tan^{-1}x}{(1 + x^2)^{3/2}} dx \] Let \( x = \tan\theta \), which implies \( dx = \sec^2\theta d\theta \) and \( \theta = \tan^{-1}x \). Let us update the limits of integration according to this substitution:
• When \( x = 0 \), \( \theta = \tan^{-1}(0) = 0 \).
• When \( x = 1 \), \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \).

Step 2: Substituting values into the integral and simplifying the integrand.
Substitute these into the integral equation: \[ I = \int_{0}^{\frac{\pi}{4}} \frac{\tan\theta \cdot \theta}{\left(1 + \tan^2\theta\right)^{3/2}} \sec^2\theta d\theta \] Using the identity \( 1 + \tan^2\theta = \sec^2\theta \): \[ \left(1 + \tan^2\theta\right)^{3/2} = \left(\sec^2\theta\right)^{3/2} = \sec^3\theta \] Now substitute this back into the expression: \[ I = \int_{0}^{\frac{\pi}{4}} \frac{\theta \tan\theta \sec^2\theta}{\sec^3\theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{\theta \tan\theta}{\sec\theta} d\theta \] Convert to sine and cosine terms: \[ \frac{\tan\theta}{\sec\theta} = \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}} = \sin\theta \quad \Rightarrow \quad I = \int_{0}^{\frac{\pi}{4}} \theta \sin\theta d\theta \]

Step 3: Integrating by parts and evaluating the final value.
Using Integration by Parts (\(\int u v d\theta = u \int v d\theta - \int (u' \int v d\theta) d\theta\)), let \( u = \theta \) and \( v = \sin\theta \): \[ I = \left[ \theta (-\cos\theta) \right]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} (1)(-\cos\theta) d\theta \] \[ I = \left[ -\theta \cos\theta \right]_{0}^{\frac{\pi}{4}} + \left[ \sin\theta \right]_{0}^{\frac{\pi}{4}} \] Apply the upper and lower limits: \[ I = \left( -\frac{\pi}{4}\cos\left(\frac{\pi}{4}\right) - 0 \right) + \left( \sin\left(\frac{\pi}{4}\right) - \sin(0) \right) \] \[ I = -\frac{\pi}{4}\left(\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\left(1 - \frac{\pi}{4}\right) = \frac{4 - \pi}{4\sqrt{2}} \]