Question
Write the following function in the simplest form:
$\tan^{-1}\frac{\sqrt{1+x^{2}}-1}{x},x\neq0$
$\tan^{-1}\frac{\sqrt{1+x^{2}}-1}{x},x\neq0$
$=\tan^{-1}\frac{\sec\theta-1}{\tan\theta}=\tan^{-1}\bigg(\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\bigg)$
$=\tan{-1}\bigg(\frac{1-\cos\theta}{\sin\theta}\bigg)=\tan^{-1}\bigg(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\bigg)$$=\tan^{-1}\bigg(\tan\frac{\theta}{2}\bigg)=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x$
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$\int\text{x}^3\cos\text{x}^2\text{dx}$
Function
$\text{y}=\sin\text{x}$