The value of int frac{sin x - cos x}{sin x + | vedclass

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Question

(D) Let $I = \int \frac{\sin x - \cos x}{\sin x + \cos x} \,dx$. Substitute $t = \sin x + \cos x$. Then,$dt = (\cos x - \sin x) \,dx$,which implies $-

Vedclass · Accepted Answer

$\log \left( \frac{1}{\sin x + \cos x} \right) + c$

Vedclass · Answer

(D) Let $I = \int \frac{\sin x - \cos x}{\sin x + \cos x} \,dx$. Substitute $t = \sin x + \cos x$. Then,$dt = (\cos x - \sin x) \,dx$,which implies $-dt = (\sin x - \cos x) \,dx$. Substituting these into the integral,we get: $I = \int \frac{-dt}{t} = -\log |t| + c$. Using the property of logarithms,$-\log |t| = \log |t^{-1}| = \log \left( \frac{1}{|t|} \right) + c$. Substituting $t = \sin x + \cos x$ back,we get: $I = \log \left( \frac{1}{|\sin x + \cos x|} \right) + c$.

The value of $\int \frac{\sin x - \cos x}{\sin x + \cos x} \,dx$ is

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