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CBSE Class 12-science Answered

pl  integrate the following. Pl guide if any fundamental clue involved in solving such problems.
1.integral sin to the power of negative 1 end exponent space left parenthesis fraction numerator 2 x over denominator 1 plus x squared end fraction right parenthesis space d x
2.integral si n to the power of negative 1 end exponent square root of fraction numerator x over denominator a plus x end fraction end root space d x
3.integral fraction numerator d x over denominator square root of s i n cubed x. c o s x end root end fraction space
4.integral cos left parenthesis 2 space c o t to the power of negative 1 end exponent square root of fraction numerator 1 minus x over denominator 1 plus x end fraction end root space right parenthesis space d x
5.integral fraction numerator x over denominator square root of 1 minus x squared plus x to the power of 4 end root end fraction space d x
6.integral fraction numerator d x over denominator x left parenthesis x to the power of 4 minus 1 right parenthesis end fraction
7.integral fraction numerator log x over denominator left parenthesis 1 plus log x right parenthesis squared end fraction d x
Asked by mishrapragu1998 | 14 Feb, 2016, 12:14: PM
answered-by-expert Expert Answer
begin mathsize 11px style The space first space step space in space solving space these space types space of space indefinite space integrals space is space to space choose space appropriate space substitution space to space simplify space the space integrals. 1. space integral sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses dx Substitute space straight x equals tanθ rightwards double arrow dx equals sec squared θdθ The space integral space becomes integral sin to the power of negative 1 end exponent open parentheses fraction numerator 2 tanθ over denominator 1 plus tan squared straight theta end fraction close parentheses sec squared θdθ equals integral sin to the power of negative 1 end exponent open parentheses sin 2 straight theta close parentheses sec squared θdθ equals 2 integral θsec squared θdθ Using space integration space by space parts 2 integral θsec squared θdθ equals 2 open square brackets θtanθ minus integral open parentheses 1 cross times tanθ close parentheses dθ close square brackets equals 2 open square brackets θtanθ minus log open vertical bar secθ close vertical bar close square brackets Resubstituting space the space value space of space tanθ comma space we space get integral sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses dx equals 2 open square brackets xtan to the power of negative 1 end exponent straight x minus log open vertical bar square root of 1 plus straight x squared end root close vertical bar close square brackets plus straight C Similarly space other space questions space can space be space solved space with space appropriate space substitutions. In space 2 nd space question comma space substitute space straight x equals atan squared straight theta In space 3 rd space question comma space substitute space tanx equals straight t In space 4 th space question comma space substitute comma space straight x equals cosθ In space 4 th space question comma space substitute comma space straight x squared equals straight t In space 5 th space question comma space substitute space straight x to the power of 4 minus 1 equals straight t In space 6 th space question comma space substitute space logx equals straight t All space these space substitutions space will space help space simplify space the space integral space and bring space it space to space some space standard space form space whose space integral space is space already space known. Hope space this space helps. end style
Answered by satyajit samal | 15 Feb, 2016, 05:47: PM

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