0-pi-4-dx-sin-x-1-1-

Question Number 182769 by cortano1 last updated on 14/Dec/22

\underset{0}{\int^{π / 4}} \frac{d x}{\sin (x - 1) - 1} = ?

Answered by ARUNG_Brandon_MBU last updated on 14/Dec/22

I = \int_{0}^{\frac{π}{4}} \frac{d x}{\sin (x - 1) - 1} = \int_{0}^{\frac{π}{4}} \frac{\sin (x - 1) + 1}{\sin^{2} (x - 1) - 1} d x

= - \int_{- 1}^{\frac{π}{4} - 1} \frac{\sin t + 1}{\cos^{2} t} d t = {[\frac{1}{\cos t} + \tan t]}_{\frac{π}{4} - 1}^{- 1}

= \sec (1) + \tan (1) - \sec (\frac{π}{4} - 1) - \tan (\frac{π}{4} - 1)