# How do you integrate int x /sqrt(1 + x^2) dx using trigonometric substitution?

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \sqrt{1 + {x}^{2}} + C$

#### Explanation:

Integrate $\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx}$ by Trigonometric substitution

Solution:

Let $x = \tan \theta$
Let ${x}^{2} = {\tan}^{2} \theta$
Let $\mathrm{dx} = {\sec}^{2} \theta \cdot d \theta$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \int \frac{\tan \theta \cdot {\sec}^{2} \theta \cdot d \theta}{\sqrt{1 + {\tan}^{2} \theta}} = \int \frac{\tan \theta \cdot {\sec}^{2} \theta \cdot d \theta}{\sqrt{{\sec}^{2} \theta}}$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \int \frac{\tan \theta \cdot {\sec}^{2} \theta \cdot d \theta}{\sec} \theta$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \int \tan \theta \cdot \sec \theta \cdot d \theta$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \int \left(\sin \frac{\theta}{\cos} \theta \cdot \frac{1}{\cos} \theta\right) \cdot d \theta$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \int \sin \frac{\theta}{\cos} ^ 2 \theta d \theta$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \int {\cos}^{- 2} \theta \cdot \sin \theta \cdot d \theta$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = - \int {\cos}^{- 2} \theta \cdot \left(- \sin \theta\right) \cdot d \theta$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = - {\left(\cos \theta\right)}^{- 2 + 1} / \left(- 2 + 1\right)$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = {\left(\cos \theta\right)}^{- 1} = \sec \theta$

Our Right Triangle with acute angle $\theta$

$\tan \theta = \frac{x}{1}$

$\sec \theta = \frac{\sqrt{1 + {x}^{2}}}{1}$

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = {\left(\cos \theta\right)}^{- 1} = \sec \theta = \frac{\sqrt{1 + {x}^{2}}}{1}$

Therefore

$\int \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \sqrt{1 + {x}^{2}} + C$

God bless....I hope the explanation is useful.