Evaluate: $\int_{0}^{4} \frac{dx}{1 + x^{2}} dx$

Solution

$\int_{0}^{4} \frac{dx}{1 + x^{2}} dx Let X = tanθ \Rightarrow θ = \tan^{- 1} X dx = \sec^{2} θ dθ When X = 0, θ = \tan^{- 1} (0) = 0 When X = 1, θ = \tan^{- 1} 1 = \frac{π}{4} ∴ \int_{0}^{4} \frac{dx}{1 + x^{2}} = \int_{0}^{\frac{π}{4}} \frac{\sec^{2} θ}{1 + \tan^{2} θ} dθ = \int_{0}^{\frac{π}{4}} \frac{\sec^{2} θ}{\sec^{2} θ} dθ = \int_{0}^{\frac{π}{4}} dθ = {[θ]}_{0}^{\frac{π}{4}} = [\frac{π}{4} - 0] = \frac{π}{4}$

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Evaluate: $\int_{0}^{4} \frac{dx}{1 + x^{2}} dx$

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Evaluate: $\int_{0}^{4} \frac{dx}{1 + x^{2}} dx$