উচ্চতর গণিত ১ম পত্র প্রশ্ন

(৩) ধরি $x=\sin^2\theta$, তাহলে $\sqrt{x}=\sin\theta$, $\sqrt{1-x}=\cos\theta$ এবং $dx=2\sin\theta\cos\theta\,d\theta$।

$I=\int\frac{2\sin\theta\cos\theta}{\sin\theta+\cos\theta}d\theta$

বিশ্লেষণ: $2\sin\theta\cos\theta=(\sin\theta+\cos\theta)^2-1$। তাই
$I=\int(\sin\theta+\cos\theta)\,d\theta-\int\frac{d\theta}{\sin\theta+\cos\theta}$

প্রথম অংশ: $\int(\sin\theta+\cos\theta)d\theta=-\cos\theta+\sin\theta$।

দ্বিতীয় অংশ: $\sin\theta+\cos\theta=\sqrt2\sin\left(\theta+\frac{\pi}{4}\right)$, তাই
$\int\frac{d\theta}{\sqrt2\sin(\theta+\frac{\pi}{4})}=\frac{1}{\sqrt2}\ln\left|\tan\left(\frac{\theta}{2}+\frac{\pi}{8}\right)\right|$

সুতরাং $I=\sin\theta-\cos\theta-\frac{1}{\sqrt2}\ln\left|\tan\left(\frac{\theta}{2}+\frac{\pi}{8}\right)\right|+C$।

$\theta=\sin^{-1}\sqrt{x}$ বসিয়ে:

$\int\frac{dx}{\sqrt x+\sqrt{1-x}}=\sqrt{x}-\sqrt{1-x}-\frac{1}{\sqrt2}\ln\left|\tan\left(\frac{1}{2}\sin^{-1}\sqrt{x}+\frac{\pi}{8}\right)\right|+C$