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CBSE Class 12 Math 2008 Solved Paper

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Question : 7 of 29
Marks: +1, -0
Evaluate: ∫04dx1+x2\int\limits_{0}^{4} \frac{dx}{1+x^2} dx
Solution:  
∫04dx1+x2\int\limits_{0}^{4} \frac{dx}{1+x^2} dx
Let x = tan θ ⇒ θ = tan⁡−1\tan^{-1} x
dx = sec⁡2\sec^2 θ d θ
When x = 0 , θ = tan⁡−1\tan^{-1} (0) = 0
When x = 1, θ = tan⁡−1\tan^{-1} 1 = π4\frac{\pi}{4}
∴ ∫04dx1+x2\int\limits_{0}^{4} \frac{dx}{1+x^2} = ∫0π4\int\limits_{0}^{\frac{\pi}{4}} sec⁡2θ1+tan⁡2θ\frac{\sec^2\theta}{1+\tan^2\theta} dθ
= ∫0π4sec⁡2θsec⁡2θ\int\limits_{0}^{\frac{\pi}{4}} \frac{\sec^2\theta}{\sec^2\theta} dθ
= ∫0π4\int\limits_{0}^{\frac{\pi}{4}} dθ
= [θ]0π4\left[\theta\right]_{0}^{\frac{\pi}{4}}
= [π4−0]\left[\frac{\pi}{4} - 0\right]
= π4\frac{\pi}{4}
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