intdx/(1+sinx) এর মান কত ? 

প্রশ্ন: ∫dx/(1+sinx) এর মান কত?

সমাধান:

আমরা জানি, sinx = 2tan(x/2) / (1 + tan²(x/2))

সুতরাং,

∫dx/(1+sinx) = ∫dx / (1 + 2tan(x/2) / (1 + tan²(x/2)))

= ∫(1 + tan²(x/2)) dx / (1 + tan²(x/2) + 2tan(x/2))

= ∫sec²(x/2) dx / (1 + tan(x/2))²

ধরি, tan(x/2) = t

তাহলে, sec²(x/2) * (1/2) dx = dt

sec²(x/2) dx = 2 dt

সুতরাং,

∫2 dt / (1+t)² = 2 ∫(1+t)⁻² dt

= 2 * (-1) * (1+t)⁻¹ + C

= -2 / (1+t) + C

= -2 / (1 + tan(x/2)) + C

এখন, -2 / (1 + tan(x/2)) = -2 / (1 + sin(x/2) / cos(x/2))

= -2cos(x/2) / (cos(x/2) + sin(x/2))

= -√2 * (√2/2 * cos(x/2) - √2/2 * sin(x/2) + √2/2 * sin(x/2) + √2/2 * cos(x/2)) / (cos(x/2) + sin(x/2))

= -√2 * [sin(π/4)cos(x/2)-cos(π/4)sin(x/2)+sin(π/4)sin(x/2)+cos(π/4)cos(x/2)] / (cos(x/2) + sin(x/2))

এখানে একটু অন্যভাবে করা যাক,

-2/(1+tan(x/2)) = \(\frac{-2}{1+\frac{sinx}{1+cosx}}\) = \(\frac{-2(1+cosx)}{1+cosx+sinx}\)

= \(\frac{-2-2cosx}{1+cosx+sinx}\) = \(\frac{-(1+cosx+sinx)-(1+cosx-sinx)}{1+cosx+sinx}\)

= \(-1-\frac{1+cosx-sinx}{1+cosx+sinx}\) = \(-1-\frac{2cos^2(x/2)-2sin(x/2)cos(x/2)}{2cos^2(x/2)+2sin(x/2)cos(x/2)}\)

= \(-1-\frac{cos(x/2)-sin(x/2)}{cos(x/2)+sin(x/2)}\) = \(-1-\frac{1-tan(x/2)}{1+tan(x/2)}\) = \(-1-tan(\pi/4-x/2)\)

আমরা জানি, tanx - secx = \(\frac{sinx-1}{cosx}\) = \(\frac{2sin(x/2)cos(x/2)-sin^2(x/2)-cos^2(x/2)}{cos^2(x/2)-sin^2(x/2)}\)

= \(\frac{-(cos(x/2)-sin(x/2))^2}{(cos(x/2)+sin(x/2))(cos(x/2)-sin(x/2))}\) = \(-\frac{cos(x/2)-sin(x/2)}{cos(x/2)+sin(x/2)}\)

সুতরাং, ∫dx/(1+sinx) = tanx - secx + C 🥳