Mathematics

Evaluate the following integral
$$\int { \cfrac { 1-\cot { x }  }{ 1+\cot { x }  }  } dx$$


SOLUTION
$$I=\displaystyle\int{\dfrac{1-\cot{x}}{1+\cot{x}}dx}$$

$$I=\displaystyle\int{\dfrac{1-\dfrac{\cos{x}}{\sin{x}}}{1+\dfrac{\cos{x}}{\sin{x}}}dx}$$

$$I=\displaystyle\int{\dfrac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}$$

Let $$t=\sin{x}+\cos{x}\Rightarrow\,dt=\left(\cos{x}-\sin{x}\right)dx$$

$$\Rightarrow\,dt=-\left(\sin{x}-\cos{x}\right)dx$$

$$I=-\displaystyle\int{\dfrac{dt}{t}}$$

$$I=-\log{\left|t\right|}+c$$

$$\therefore\,I=-\log{\left|\sin{x}+\cos{x}\right|}+c$$ where $$t=\sin{x}+\cos{x}$$

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Subjective Medium Published on 17th 09, 2020
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