Question

Evaluate ${\int }_0^{\pi /4}\frac{\cos x-\sin x}{10+\sin 2x}dx$

• A. $\frac{1}{3}({\tan }^{-1}\frac{\sqrt{2}}{3}+{\tan }^{-1}\frac{1}{3})$
• B. $\frac{1}{3}({\tan }^{-1}\frac{\sqrt{1}}{3}-{\cot }^{-1}\frac{2}{3})$
• C. $\frac{1}{3}({\tan }^{-1}\frac{\sqrt{2}}{3}-{\tan }^{-1}\frac{1}{3})$
• D. $\frac{1}{3}({\tan }^{-1}\frac{\sqrt{1}}{3}-{\cot }^{-1}\frac{1}{3})$

Answer 1

Answer: (C)

$\frac{1}{3}({\tan }^{-1}\frac{\sqrt{2}}{3}-{\tan }^{-1}\frac{1}{3})$

Let $I={\int }_0^{\frac{\pi }{4}}\frac{\cos x-\sin x}{10+\sin 2x}dx$
Put $\cos x+\sin x=t\Rightarrow (-\sin x+\cos x)dx=dt$
$I={\int }_1^{\sqrt{2}}\frac{1}{10+(t^2-1)}dt={\int }_1^{\sqrt{2}}\frac{1}{t^2+9}dt$
$=\frac{1}{3}{\left[{\tan }^{-1}\frac{t}{3}\right]}_1^{\sqrt{2}}=\frac{1}{3}\left({\tan }^{-1}\frac{\sqrt{2}}{3}-{\tan }^{-1}\frac{1}{3}\right)$