অংশায়ন পদ্ধতিতে যোগজীকরণ (Integration by parts)

অনুশীলনী \(10.F / Q.2\)-এর প্রশ্নসমুহ

যোজিত ফল নির্ণয় করঃ
\(Q.2.(i)\) \(\int{\cos^{-1}{x}dx}\)
উত্তরঃ \(x\cos^{-1}{x}-\sqrt{1-x^2}+c\)
[ রাঃ২০১০; যঃ২০০৮; চঃ২০০৭; কুঃ২০০৫ ]

\(Q.2.(ii)\) \(\int{\sin^{-1}{x}dx}\)
উত্তরঃ \(x\sin^{-1}{x}+\sqrt{1-x^2}+c\)
[ যঃ২০১০; সিঃ২০০৩ ]

\(Q.2.(iii)\) \(\int{x\tan^{-1}{x}dx}\)
উত্তরঃ \(\frac{1}{2}(x^2+1)\tan^{-1}{x}-\frac{1}{2}x+c\)
[ দিঃ২০১১; কুঃ২০১০; সিঃ২০০৮,২০০৪; রাঃ২০০৬; যঃ২০০৬,২০০১; বঃ২০০২; মাঃ২০০৮ ]

\(Q.2.(iv)\) \(\int{x\sin^{-1}{x}dx}\)
উত্তরঃ \(\frac{1}{2}x^2\sin^{-1}{x}+\frac{1}{4}x\sqrt{1-x^2}-\frac{1}{4}\sin^{-1}{x}+c\)
[ ঢাঃ২০০৭; মাঃ২০০৬,২০০৩,২০০২ ]

\(Q.2.(v)\) \(\int{x \ cosec^{-1}{\frac{1}{x}}dx}\)
উত্তরঃ \(\frac{1}{2}x^2\sin^{-1}{x}+\frac{1}{4}x\sqrt{1-x^2}-\frac{1}{4}\sin^{-1}{x}+c\)
[ চঃ২০১৭ ]

\(Q.2.(vi)\) \(\int{mx\sin^{-1}{x}dx}\)
উত্তরঃ \(m\left(\frac{1}{2}x^2\sin^{-1}{x}+\frac{1}{4}x\sqrt{1-x^2}-\frac{1}{4}\sin^{-1}{x}\right)+c\)
[ দিঃ২০১৭ ]

\(Q.2.(vii)\) \(\int{\tan^{-1}{x}dx}\)
উত্তরঃ \(x\tan^{-1}{x}-\frac{1}{2}\ln{|1+x^2|}+c\)
[ ঢাঃ২০০৪; কুঃ২০০২ ]

\(Q.2.(viii)\) \(\int{\tan^{-1}{\frac{x}{5}}dx}\)
উত্তরঃ \(x\tan^{-1}{\frac{x}{5}}-\frac{5}{2}\ln{|x^2+25|}+c\)
[ বঃ২০১৭ ]

\(Q.2.(ix)\) \(\int{e^{2x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{5}e^{2x}(2\cos{x}+\sin{x})+c\)

\(Q.2.(x)\) \(\int{e^{2x}\sin{x}dx}\)
উত্তরঃ \(\frac{1}{5}e^{2x}(2\sin{x}-\cos{x})+c\)
[ সিঃ২০০২ ]

\(Q.2.(xi)\) \(\int{e^{x}\sin{2x}dx}\)
উত্তরঃ \(\frac{1}{5}e^{x}(\sin{2x}-2\cos{2x})+c\)
[ সিঃ২০১০; রাঃ,দিঃ২০০৯ ]

\(Q.2.(xii)\) \(\int{e^{x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{2}e^{x}(\cos{x}+\sin{x})+c\)
[ ঢাঃ২০১০; মাঃ২০১৩; বুয়েটঃ২০০৪,২০০৬ ]

\(Q.2.(xiii)\) \(\int{e^{x}\sin{x}dx}\)
উত্তরঃ \(\frac{1}{2}e^{x}(\sin{x}-\cos{x})+c\)
[ দিঃ,রাঃ২০১৪; কুঃ২০১৩,২০০৮; মাঃ২০০৯; ঢাঃ২০০৮,২০০৩ ]

\(Q.2.(xiv)\) \(\int{e^{ax}\sin{bx}dx}\)
উত্তরঃ \(\frac{e^{ax}(a\cos{bx}+b\sin{bx})}{a^2+b^2}+c\)
[ রাঃ২০০৬; ঢাঃ২০০৫ ]

\(Q.2.(xv)\) \(\int{x\cos^2{x}dx}\)
উত্তরঃ \(\frac{1}{8}(2x^2+2x\sin{2x}+\cos{2x})+c\)
[ সিঃ২০০৭ ]

\(Q.2.(xvi)\) \(\int{x\sin^2{x}dx}\)
উত্তরঃ \(\frac{1}{8}(2x^2-2x\sin{2x}-\cos{2x})+c\)
[ কুয়েটঃ২০০৫-২০০৬ ]
\(Q.2.(xvii)\) \(\int{x\sin{x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{8}(\sin{2x}-2x\cos{2x})+c\)

\(Q.2.(xviii)\) \(\int{x\sin^3{x}dx}\)
উত্তরঃ \(\frac{1}{36}(27\sin{x}+3x\cos{3x}-27x\cos{x}-\sin{3x})+c\)

\(Q.2.(xix)\) \(\int{x\sin^2{\left(\frac{x}{2}\right)}dx}\)
উত্তরঃ \(\frac{1}{4}(x^2-2x\sin{2x}-2\cos{2x})+c\)

\(Q.2.(xx)\) \(\int{x\cos{2x}\cos{3x}dx}\)
উত্তরঃ \(\frac{1}{50}(25x\sin{x}+25\cos{x}+5x\sin{5x}+\cos{5x})+c\)

\(Q.2.(xxi)\) \(\int{x\sin{x}\sin{2x}dx}\)
উত্তরঃ \(\frac{1}{18}(9x\sin{x}+9\cos{x}-3x\sin{3x}-\cos{3x})+c\)

\(Q.2.(xxii)\) \(\int{x^2\cos^2{\frac{x}{2}}dx}\)
উত্তরঃ \(\frac{1}{6}(x^3+3x^2\sin{x}+6x\cos{x}-6\sin{x})+c\)

\(Q.2.(xxiii)\) \(\int{x\tan^2{x}dx}\)
উত্তরঃ \(x\tan{x}+\ln{|\cos{x}|}-\frac{1}{2}x^2+c\)
[ রাঃ২০০৫; সিঃ২০০৫; ঢাঃ২০০২]

\(Q.2.(xxiv)\) \(\int{x\sec^2{x}dx}\)
উত্তরঃ \(x\tan{x}+\ln{|\cos{x}|}+c\)
[ চঃ২০১৪ ]

\(Q.2.(xxv)\) \(\int{x\tan^{-1}{(x^2)}dx}\)
উত্তরঃ \(\frac{1}{4}(2x^2\tan^{-1}{x^2}-\ln{|1+x^4|})+c\)

\(Q.2.(xxvi)\) \(\int{x\sin^{-1}{(x^2)}dx}\)
উত্তরঃ \(\frac{1}{2}(x^2\sin^{-1}{(x^2)}+\sqrt{1-x^4})+c\)

\(Q.2.(xxvii)\) \(\int{\cos^{-1}{\left(\frac{1-x^2}{1+x^2}\right)}dx}\)
উত্তরঃ \(2x\tan^{-1}{x}-\ln{|(1+x^2)|}+c\)

\(Q.2.(xxviii)\) \(\int{\sin^{-1}{\left(\frac{2x}{1+x^2}\right)}dx}\)
উত্তরঃ \(2x\tan^{-1}{x}-\frac{1}{2}\ln{|(1+x^2)|}+c\)

\(Q.2.(xxix)\) \(\int{\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}dx}\)
উত্তরঃ \(2x\tan^{-1}{x}-\frac{1}{2}\ln{|(1+x^2)|}+c\)

\(Q.2.(xxx)\) \(\int{\sin^{-1}{\sqrt{\frac{x}{a+x}}}dx}\)
উত্তরঃ \((a+x)\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\sqrt{(ax)}+c\)
[ বুয়েটঃ২০১১-২০১২ ]

\(Q.2.(xxxi)\) \(\int{\tan^{-1}{\sqrt{\frac{1-x}{1+x}}}dx}\)
উত্তরঃ \(\frac{1}{2}(x\cos^{-1}{x}-\sqrt{1-x^2})+c\)

\(Q.2.(xxxii)\) \(\int{\cos^{-1}{\sqrt{x}}dx}\)
উত্তরঃ \(x\cos^{-1}{\sqrt{x}}+\frac{1}{2}\sin^{-1}{\sqrt{x}}-\frac{\sqrt{x(1-x)}}{2}+c\)

\(Q.2.(xxxiii)\) \(\int{\sin^{-1}{\sqrt{x}}dx}\)
উত্তরঃ \(\left(x-\frac{1}{2}\right)\sin^{-1}{\sqrt{x}}+\frac{1}{2}\sqrt{x(1-x)}+c\)

\(Q.2.(xxxiv)\) \(\int{\tan^{-1}{\sqrt{x}}dx}\)
উত্তরঃ \(\left(x+1\right)\tan^{-1}{\sqrt{x}}-\sqrt{x}+c\)

\(Q.2.(xxxv)\) \(\int{\sec^{-1}{x}dx}\)
উত্তরঃ \(x\sec^{-1}{x}-\ln{|x+\sqrt{x^2-1}|}+c\)
1 2 3 4 5