AI-তৈরি নমুনা উত্তর
ধরি, I = ∫ d x x + 1 − x I = \int \frac{dx}{\sqrt{x} + \sqrt{1-x}} I = ∫ x + 1 − x d x
হর ও লবকে ( x − 1 − x ) (\sqrt{x} - \sqrt{1-x}) ( x − 1 − x ) I = ∫ x − 1 − x ( x + 1 − x ) ( x − 1 − x ) d x = ∫ x − 1 − x 2 x − 1 d x I = \int \frac{\sqrt{x} - \sqrt{1-x}}{(\sqrt{x} + \sqrt{1-x})(\sqrt{x} - \sqrt{1-x})} dx = \int \frac{\sqrt{x} - \sqrt{1-x}}{2x - 1} dx I = ∫ ( x + 1 − x ) ( x − 1 − x ) x − 1 − x d x = ∫ 2 x − 1 x − 1 − x d x
ধরি, x = sin 2 θ x = \sin^2\theta x = sin 2 θ d x = 2 sin θ cos θ d θ = sin 2 θ d θ dx = 2\sin\theta\cos\theta\, d\theta = \sin 2\theta\, d\theta d x = 2 sin θ cos θ d θ = sin 2 θ d θ 2 x − 1 = 2 sin 2 θ − 1 = − cos 2 θ 2x - 1 = 2\sin^2\theta - 1 = -\cos 2\theta 2 x − 1 = 2 sin 2 θ − 1 = − cos 2 θ
সুতরাং,I = ∫ sin θ − cos θ − cos 2 θ ( 2 sin θ cos θ ) d θ I = \int \frac{\sin\theta - \cos\theta}{-\cos 2\theta} (2\sin\theta\cos\theta)\, d\theta I = ∫ − c o s 2 θ s i n θ − c o s θ ( 2 sin θ cos θ ) d θ = ∫ cos θ − sin θ cos 2 θ − sin 2 θ ( 2 sin θ cos θ ) d θ = \int \frac{\cos\theta - \sin\theta}{\cos^2\theta - \sin^2\theta} (2\sin\theta\cos\theta)\, d\theta = ∫ c o s 2 θ − s i n 2 θ c o s θ − s i n θ ( 2 sin θ cos θ ) d θ = ∫ 2 sin θ cos θ cos θ + sin θ d θ = \int \frac{2\sin\theta\cos\theta}{\cos\theta + \sin\theta} d\theta = ∫ c o s θ + s i n θ 2 s i n θ c o s θ d θ = ∫ ( cos θ + sin θ ) 2 − 1 cos θ + sin θ d θ = \int \frac{(\cos\theta + \sin\theta)^2 - 1}{\cos\theta + \sin\theta} d\theta = ∫ c o s θ + s i n θ ( c o s θ + s i n θ ) 2 − 1 d θ = ∫ ( cos θ + sin θ ) d θ − ∫ d θ 2 cos ( θ − π 4 ) = \int (\cos\theta + \sin\theta) d\theta - \int \frac{d\theta}{\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)} = ∫ ( cos θ + sin θ ) d θ − ∫ 2 c o s ( θ − 4 π ) d θ = sin θ − cos θ − 1 2 ∫ sec ( θ − π 4 ) d θ = \sin\theta - \cos\theta - \frac{1}{\sqrt{2}} \int \sec\left(\theta - \frac{\pi}{4}\right) d\theta = sin θ − cos θ − 2 1 ∫ sec ( θ − 4 π ) d θ = sin θ − cos θ − 1 2 ln ∣ sec ( θ − π 4 ) + tan ( θ − π 4 ) ∣ + C = \sin\theta - \cos\theta - \frac{1}{\sqrt{2}} \ln \left| \sec\left(\theta - \frac{\pi}{4}\right) + \tan\left(\theta - \frac{\pi}{4}\right) \right| + C = sin θ − cos θ − 2 1 ln sec ( θ − 4 π ) + tan ( θ − 4 π ) + C
যেহেতু sin θ = x \sin\theta = \sqrt{x} sin θ = x cos θ = 1 − x \cos\theta = \sqrt{1-x} cos θ = 1 − x tan ( θ − π 4 ) = tan θ − 1 tan θ + 1 = x − 1 − x x + 1 − x \tan\left(\theta - \frac{\pi}{4}\right) = \frac{\tan\theta - 1}{\tan\theta + 1} = \frac{\sqrt{x} - \sqrt{1-x}}{\sqrt{x} + \sqrt{1-x}} tan ( θ − 4 π ) = t a n θ + 1 t a n θ − 1 = x + 1 − x x − 1 − x sec ( θ − π 4 ) = 2 cos θ + sin θ = 2 x + 1 − x \sec\left(\theta - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{\cos\theta + \sin\theta} = \frac{\sqrt{2}}{\sqrt{x} + \sqrt{1-x}} sec ( θ − 4 π ) = c o s θ + s i n θ 2 = x + 1 − x 2
অতএব,I = x − 1 − x − 1 2 ln ∣ 2 + x − 1 − x x + 1 − x ∣ + C I = \sqrt{x} - \sqrt{1-x} - \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2} + \sqrt{x} - \sqrt{1-x}}{\sqrt{x} + \sqrt{1-x}} \right| + C I = x − 1 − x − 2 1 ln x + 1 − x 2 + x − 1 − x + C