Question

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

• A. $\frac{1}{\sin x-\cos x}+C$
• B. $\frac{-1}{\sin x-\cos x}+C$
• C. $\frac{-1}{\sin x+\cos x}+C$
• D. None of these

Answer 1

Answer: (B)

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

$\int \frac{\sin x+\cos x}{(\sin x-\cos x{)}^2}dx$

Let $\sin x-\cos x=t$

$\frac{d}{dx}(\sin x-\cos x)=\frac{dt}{dx}$

$(\cos x+\sin x)dx=dt$

$\int \frac{dt}{t^2}$

$\int t^{-2}dt$

$=-t^{-1}+C=\frac{-1}{\sin x-\cos x}+C$

$\frac{d}{dx}\left(\sin x\right)=\cos x$

$\frac{d}{dx}\left(\cos x\right)=\sin x$

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$