Mathematics

Evaluate
$$\displaystyle\int {\dfrac{{\sin x + \cos x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}dx} $$


ANSWER

$$\dfrac {-1}{\sin x-\cos x}+C$$


SOLUTION
$$\int \dfrac{\sin x + \cos x}{(\sin x - \cos x)^2} dx$$
Let $$\sin x - \cos x = t$$
$$\dfrac{d}{dx} (\sin x - \cos x) = \dfrac{dt}{dx}$$
$$(\cos x + \sin x)dx = dt$$
$$\int \dfrac{dt}{t^2}$$
$$\int t^{-2} dt$$
$$= -t^{-1} + C = \dfrac{-1}{\sin x - \cos x} + C$$
$$\dfrac{d}{dx} (\sin x) = \cos x$$
$$\dfrac{d}{dx} (\cos x) = \sin x$$
$$\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$$
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Single Correct Medium Published on 17th 09, 2020
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