AbdulRehman
what will be the
integration of 1/(1+x²)½ along x?



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AtootyS
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\[\int\limits 1/(1+x ^{2})^{1/2}\]
it will be \[\ln (1 + x ^{2})^{1/2}\]

AtootyS
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you know that when the numerator is the derivative of the dominator then the integration is ln "dominator"
:)

uzma
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but the derivative of denominator should be there in the numerator

uzma
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\[x= \tan \theta\]

AtootyS
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ok , the derivative of (x^2 )^1/2 is 1/2 (2x)
so 2 cancel the other 2 :)

uzma
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and for natural log, exponent should be 1

uzma
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\[\int\limits 1/x= \ln x\]

uzma
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if the exponent is other than 1, we use power rule

AbdulRehman
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yeah but we can't multiply it with 2x so ln(1+x²)½ is not the right answer

uzma
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\[d x= \sec^2 \theta d \theta\]

AtootyS
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i think its correct ..
anyways good luck ..

uzma
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\[\int\limits \sec^2 \theta/(\sqrt{1+\tan^2 \theta} )\]

uzma
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\[\int\limits \sec^2\theta/\sec \theta d \theta\]

uzma
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\[\int\limits \sec \theta d \theta\]

uzma
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\[\ln (\sec \theta + \tan \theta) +C\]

uzma
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by backward substitution
\[\tan \theta =x\]

AbdulRehman
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yeah that will give us integration of secθ that is ln(secθ+tanθ) and by substituting and we will get probably get ln(sec(tan^x)+x)

uzma
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\[\sec \theta =\sqrt{1+x^2}\]

AbdulRehman
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thanks for the help i have some questions on multiple integral usage in finding area and volume. Anyone please help me

uzma
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post it as a new Q,some one would surly look at it :)

AbdulRehman
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Uzma are you from Pakistan?

AtootyS
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ahaa ! ok waiting ur new Qs :)

uzma
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yes :)

AbdulRehman
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Can you please tell me your qualification and city

uzma
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what would u do with that :)

AbdulRehman
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nothing just to guess the level of your understanding mathematics

uzma
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mathematics is as vast, that one can hardly grasp even one area