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Question

$$\int{\dfrac{1\ dx}{1+cot x}}$$


Solution

$$\int \dfrac{1}{1+ \cot x} dx$$
$$=\int \dfrac{1}{1+ \dfrac{\cos x}{\sin x}}= \dfrac{1}{2} \int \dfrac{2 \sin x}{\sin x+ \cos x} dx =\dfrac{1}{2} \int \dfrac{(\sin x+ \cos x)-(\cos x- \sin x)}{(\sin x+ \cos x)} dx$$
$$=\dfrac{1}{2} \left[ \int 1 \ dx -\int \dfrac{\cos x- \sin x }{\sin x+ \cos x} dx\right]$$
$$\sin x+ \cos x =t$$
$$\cos x + (- \sin x) dx =dt$$
$$(\cos x- \sin ) dx = dt$$
$$\dfrac{1}{2} \left[ x- \int \dfrac{dt}{t} \right]$$
$$\dfrac{1}{2} [x- In\ |t|]+c$$
$$\dfrac{1}{2} [x- In (\sin x+ \cos x)] +c$$
$$\dfrac{x}{2} - \dfrac{1}{2} In\ (\sin x+ \cos x)+c$$
$$\therefore \int \dfrac{dx}{1+ \cot x} =\dfrac{x}{2} -\dfrac{1}{2} In (\sin x+ \cos x)+c$$

Mathematics

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