Question

Evaluate: \[ \int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) \, dx. \]

Hint:

To evaluate integrals involving trigonometric functions and inverse trigonometric functions, simplify using substitutions and apply integration by parts where necessary.

Answer 1

Step 1: Simplify \(\sin 2x\). 
Using the identity \(\sin 2x = 2 \sin x \cos x\), rewrite the integral: \[ \int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) \, dx = \int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \tan^{-1}(\sin x) \, dx. \] 

Step 2: Substitution. Let \(t = \sin x\). Then: \[ \cos x \, dx = dt, \quad \text{and} \quad \sin x = t. \] The limits of integration change as follows: - When \(x = 0\), \(\sin x = 0 \implies t = 0\), 

When \(x = \frac{\pi}{2}\), \(\sin x = 1 \implies t = 1\). The integral becomes: \[ \int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \tan^{-1}(\sin x) \, dx = \int_{0}^{1} 2t \tan^{-1}(t) \, dt. \] 

Step 3: Integration by parts. To evaluate \(\int 2t \tan^{-1}(t) \, dt\), use integration by parts: \[ \text{Let } u = \tan^{-1}(t), \quad dv = 2t \, dt. \] Then: \[ du = \frac{1}{1 + t^2} \, dt, \quad v = t^2. \] 

Using the formula for integration by parts \(\int u \, dv = uv - \int v \, du\): \[ \int 2t \tan^{-1}(t) \, dt = t^2 \tan^{-1}(t) - \int t^2 \cdot \frac{1}{1 + t^2} \, dt. \] 

Step 4: Simplify the remaining integral. Simplify \(\int \frac{t^2}{1 + t^2} \, dt\): \[ \frac{t^2}{1 + t^2} = 1 - \frac{1}{1 + t^2}. \] Thus: \[ \int \frac{t^2}{1 + t^2} \, dt = \int 1 \, dt - \int \frac{1}{1 + t^2} \, dt. \] 

Evaluate each term: \[ \int 1 \, dt = t, \quad \int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t). \] Substitute back: \[ \int \frac{t^2}{1 + t^2} \, dt = t - \tan^{-1}(t). \] 

Step 5: Final expression. 
Substitute back into the integral: \[ \int 2t \tan^{-1}(t) \, dt = t^2 \tan^{-1}(t) - \left(t - \tan^{-1}(t)\right). \] Simplify: \[ \int 2t \tan^{-1}(t) \, dt = t^2 \tan^{-1}(t) - t + \tan^{-1}(t). \] Step 6: Apply limits of integration. 

Evaluate from \(t = 0\) to \(t = 1\): \[ \int_{0}^{1} 2t \tan^{-1}(t) \, dt = \left[ t^2 \tan^{-1}(t) - t + \tan^{-1}(t) \right]_{0}^{1}. \] At \(t = 1\): \[ 1^2 \tan^{-1}(1) - 1 + \tan^{-1}(1) = 1 \cdot \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1. \] At \(t = 0\): \[ 0^2 \tan^{-1}(0) - 0 + \tan^{-1}(0) = 0. \] Thus: \[ \int_{0}^{1} 2t \tan^{-1}(t) \, dt = \frac{\pi}{2} - 1 - 0 = \frac{\pi}{2} - 1. \] 

Final Answer: \[ \int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) \, dx = \frac{\pi}{2} - 1. \]