গণিত প্রশ্ন

AI-তৈরি নমুনা উত্তর

ধরি,
$I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{dx}{1 + \sin x - \cos x}$

আমরা জানি,
$1 - \cos x = 2\sin^2\frac{x}{2}$ এবং $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$

$\therefore 1 + \sin x - \cos x = 2\sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = 2\cos^2\frac{x}{2}\left(\tan^2\frac{x}{2} + \tan\frac{x}{2}\right)$

সুতরাং,
$I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\frac{1}{2}\sec^2\frac{x}{2}}{\tan\frac{x}{2}(1 + \tan\frac{x}{2})} dx$

ধরি, $t = \tan\frac{x}{2} \implies dt = \frac{1}{2}\sec^2\frac{x}{2}\,dx$

সীমা পরিবর্তন:
$x = \frac{\pi}{3} \implies t = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$
$x = \frac{\pi}{2} \implies t = \tan\frac{\pi}{4} = 1$

$\therefore I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{dt}{t(1+t)} = \int_{\frac{1}{\sqrt{3}}}^{1} \left(\frac{1}{t} - \frac{1}{1+t}\right) dt$
$= \left[ \ln t - \ln(1+t) \right]_{\frac{1}{\sqrt{3}}}^{1}$
$= \left[ \ln\left(\frac{t}{1+t}\right) \right]_{\frac{1}{\sqrt{3}}}^{1}$
$= \ln\left(\frac{1}{1+1}\right) - \ln\left(\frac{\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\right)$
$= \ln\left(\frac{1}{2}\right) - \ln\left(\frac{1}{\sqrt{3}+1}\right)$
$= \ln\left(\frac{\sqrt{3}+1}{2}\right)$

নির্ণেয় মান: $\ln\left(\frac{\sqrt{3}+1}{2}\right)$