Question

Solve:

$\underset{0}{\overset{\pi /6}{\int }}\frac{\cos 2x}{{\left(\cos x-\sin x\right)}^2}dx$

• A. $-\log \left(\frac{\sqrt{3}-1}{2}\right)$
• B. $-\log \left(\frac{\sqrt{3}+1}{2}\right)$
• C. $\log \left(\frac{\sqrt{3}+1}{2}\right)$
• D. None of these

Answer 1

Answer: (A)

$-\log \left(\frac{\sqrt{3}-1}{2}\right)$

${\int }_0^{\pi /6}\frac{\cos 2x}{{(\cos x-\sin x)}^2}dx{\int }_0^{\pi /6}\frac{{\cos }^{2}x-{\sin }^{2}x}{{(\cos x-\sin x)}^2}dx{\int }_0^{\pi /6}\frac{(\cos x-\sin x)(\cos x+\sin x)}{{(\cos x-\sin x)}^2}dx{\int }_0^{\pi /6}\frac{(\cos x+\sin x)}{(\cos x-\sin x)}dxputt=\cos x-\sin xHence,dt=-(\cos x+\sin x)dx-{\int }_1^{\frac{\sqrt{3}-1}{2}}\frac{dt}{t}-[\log t{]}_1^{\frac{\sqrt{3}-1}{2}}-\left[\log \left(\frac{\sqrt{3}-1}{2}\right)-\log \left(1\right)\right]-\log \left(\frac{\sqrt{3}-1}{2}\right)\because \log \left(1\right)=0$