ক) AI-তৈরি নমুনা উত্তর
ধরি, I = ∫ tan − 1 x d x = ∫ 1 ⋅ tan − 1 x d x I = \int \tan^{-1} x \, dx = \int 1 \cdot \tan^{-1} x \, dx I = ∫ tan − 1 x d x = ∫ 1 ⋅ tan − 1 x d x
আংশিক সমাকলন সূত্র ব্যবহার করে,I = tan − 1 x ∫ 1 d x − ∫ ( d d x ( tan − 1 x ) ∫ 1 d x ) d x I = \tan^{-1} x \int 1 \, dx - \int \left( \frac{d}{dx}(\tan^{-1} x) \int 1 \, dx \right) dx I = tan − 1 x ∫ 1 d x − ∫ ( d x d ( tan − 1 x ) ∫ 1 d x ) d x = x tan − 1 x − ∫ x 1 + x 2 d x = x \tan^{-1} x - \int \frac{x}{1+x^2} \, dx = x tan − 1 x − ∫ 1 + x 2 x d x = x tan − 1 x − 1 2 ∫ 2 x 1 + x 2 d x = x \tan^{-1} x - \frac{1}{2} \int \frac{2x}{1+x^2} \, dx = x tan − 1 x − 2 1 ∫ 1 + x 2 2 x d x = x tan − 1 x − 1 2 ln ( 1 + x 2 ) + c = x \tan^{-1} x - \frac{1}{2} \ln(1+x^2) + c = x tan − 1 x − 2 1 ln ( 1 + x 2 ) + c c c c
উত্তর: x tan − 1 x − 1 2 ln ( 1 + x 2 ) + c x \tan^{-1} x - \frac{1}{2} \ln(1+x^2) + c x tan − 1 x − 2 1 ln ( 1 + x 2 ) + c
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, f ( x ) = sin x f(x) = \sin x f ( x ) = sin x ∴ f ( π 2 − x ) = sin ( π 2 − x ) = cos x \therefore f\left(\frac{\pi}{2} - x\right) = \sin\left(\frac{\pi}{2} - x\right) = \cos x ∴ f ( 2 π − x ) = sin ( 2 π − x ) = cos x
প্রদত্ত সমাকল,I = ∫ 0 π 2 cos x 9 − sin 2 x d x I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{9 - \sin^2 x} \, dx I = ∫ 0 2 π 9 − s i n 2 x c o s x d x
ধরি, u = sin x u = \sin x u = sin x ∴ d u = cos x d x \therefore du = \cos x \, dx ∴ d u = cos x d x x = 0 x = 0 x = 0 u = sin 0 = 0 u = \sin 0 = 0 u = sin 0 = 0 x = π 2 x = \frac{\pi}{2} x = 2 π u = sin ( π 2 ) = 1 u = \sin\left(\frac{\pi}{2}\right) = 1 u = sin ( 2 π ) = 1
∴ I = ∫ 0 1 d u 3 2 − u 2 \therefore I = \int_{0}^{1} \frac{du}{3^2 - u^2} ∴ I = ∫ 0 1 3 2 − u 2 d u = [ 1 2 ⋅ 3 ln ∣ 3 + u 3 − u ∣ ] 0 1 = \left[ \frac{1}{2 \cdot 3} \ln\left| \frac{3+u}{3-u} \right| \right]_{0}^{1} = [ 2 ⋅ 3 1 ln 3 − u 3 + u ] 0 1 = 1 6 [ ln ( 4 2 ) − ln ( 3 3 ) ] = \frac{1}{6} \left[ \ln\left(\frac{4}{2}\right) - \ln\left(\frac{3}{3}\right) \right] = 6 1 [ ln ( 2 4 ) − ln ( 3 3 ) ] = 1 6 ( ln 2 − ln 1 ) = \frac{1}{6} (\ln 2 - \ln 1) = 6 1 ( ln 2 − ln 1 ) = 1 6 ln 2 = \frac{1}{6} \ln 2 = 6 1 ln 2
উত্তর: 1 6 ln 2 \frac{1}{6} \ln 2 6 1 ln 2
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে,g ( x , y ) = 25 x 2 + 36 y 2 − 900 = 0 g(x, y) = 25x^2 + 36y^2 - 900 = 0 g ( x , y ) = 25 x 2 + 36 y 2 − 900 = 0 ⇒ x 2 36 + y 2 25 = 1 \Rightarrow \frac{x^2}{36} + \frac{y^2}{25} = 1 ⇒ 36 x 2 + 25 y 2 = 1 ⇒ x 2 6 2 + y 2 5 2 = 1 \Rightarrow \frac{x^2}{6^2} + \frac{y^2}{5^2} = 1 ⇒ 6 2 x 2 + 5 2 y 2 = 1 x x x a = 6 , b = 5 a = 6, b = 5 a = 6 , b = 5
এখানে, y = ± 5 6 36 − x 2 y = \pm \frac{5}{6} \sqrt{36 - x^2} y = ± 6 5 36 − x 2
উপবৃত্তটি এবং x = 3 x = 3 x = 3 x = 3 x = 3 x = 3 x = 6 x = 6 x = 6 x x x
অতএব, নির্ণেয় আবদ্ধ ক্ষুদ্রতর অংশের ক্ষেত্রফল:A = 2 ∫ 3 6 y d x A = 2 \int_{3}^{6} y \, dx A = 2 ∫ 3 6 y d x = 2 ∫ 3 6 5 6 36 − x 2 d x = 2 \int_{3}^{6} \frac{5}{6} \sqrt{36 - x^2} \, dx = 2 ∫ 3 6 6 5 36 − x 2 d x = 5 3 ∫ 3 6 6 2 − x 2 d x = \frac{5}{3} \int_{3}^{6} \sqrt{6^2 - x^2} \, dx = 3 5 ∫ 3 6 6 2 − x 2 d x = 5 3 [ x 2 36 − x 2 + 36 2 sin − 1 ( x 6 ) ] 3 6 = \frac{5}{3} \left[ \frac{x}{2} \sqrt{36 - x^2} + \frac{36}{2} \sin^{-1}\left(\frac{x}{6}\right) \right]_{3}^{6} = 3 5 [ 2 x 36 − x 2 + 2 36 sin − 1 ( 6 x ) ] 3 6 = 5 3 [ ( 0 + 18 sin − 1 ( 1 ) ) − ( 3 2 36 − 9 + 18 sin − 1 ( 1 2 ) ) ] = \frac{5}{3} \left[ \left(0 + 18 \sin^{-1}(1)\right) - \left(\frac{3}{2} \sqrt{36 - 9} + 18 \sin^{-1}\left(\frac{1}{2}\right)\right) \right] = 3 5 [ ( 0 + 18 sin − 1 ( 1 ) ) − ( 2 3 36 − 9 + 18 sin − 1 ( 2 1 ) ) ] = 5 3 [ 18 ⋅ π 2 − 3 2 27 − 18 ⋅ π 6 ] = \frac{5}{3} \left[ 18 \cdot \frac{\pi}{2} - \frac{3}{2} \sqrt{27} - 18 \cdot \frac{\pi}{6} \right] = 3 5 [ 18 ⋅ 2 π − 2 3 27 − 18 ⋅ 6 π ] = 5 3 [ 9 π − 9 3 2 − 3 π ] = \frac{5}{3} \left[ 9\pi - \frac{9\sqrt{3}}{2} - 3\pi \right] = 3 5 [ 9 π − 2 9 3 − 3 π ] = 5 3 [ 6 π − 9 3 2 ] = \frac{5}{3} \left[ 6\pi - \frac{9\sqrt{3}}{2} \right] = 3 5 [ 6 π − 2 9 3 ] = 10 π − 15 3 2 = 10\pi - \frac{15\sqrt{3}}{2} = 10 π − 2 15 3
উত্তর: 10 π − 15 3 2 10\pi - \frac{15\sqrt{3}}{2} 10 π − 2 15 3