Question

Hello, have couple integral problems that need to be answered using U-substitution method, can someone help me how to answer them step by step? my final is tomorrow ...

Answer 1

1) \[\int\limits_{}^{}4x \sec (x ^{2})\tan (x^{2})e ^{\sec (x ^{2})} dx\]

Answer 2

why dont you let u-sec(x^2) and see what u get

Answer 3

u=sec(x^2)

Answer 4

ok, let me try to answer it when u=sec(x^2)

Answer 5

\[u=\sec(x ^{2})\]
\[du=(\sec x) (\tan x)(2x)dx\]

Answer 6

should i multiply both sides by 2x?

Answer 7

you got the derivative wrong a bit , you should have sec(x^2)*tan(x^2)*2x

Answer 8

no you should not multiply anything

Answer 9

so what should i do with 4x?

Answer 10

since 4 is constant you can write 2*2 and pull out of the integral one of the 2 so you get
\[2\int\limits_{}^{}{2xsec(x^2)\tan(x^2)e^{\sec(x^2)}}\]

Answer 11

\[=> \int\limits_{}^{}4x e ^{u} du\]
i guess this is how it look like before taking the antiderivative and substitution

Answer 12

so now you can substitude for du and u where \[du=2xsec(x^2)*\tan(x^2)\]

Answer 13

oh yea, forgot about the 2x before the sec and tan,
\[=> 2\int\limits_{}^{}e ^{u} du\]
\[=> e ^{\sec x ^{2}} + c\]

is that correct?

Answer 14

you forgot the 2 , \[2*e^{\sec(x^2)}+\]

Answer 15

yea, sorry about that, i am really tired of this lol i am still 5 problems away to get some rest :p

Answer 16

you'll make it ;D http://www.wolframalpha.com/ this site can help you in future , that can solve pretty much anything :)

Answer 17

what about this one:
\[\int\limits_{}^{}\frac{ 2x-1 }{ \sqrt{3x ^{2}-3x} }\]
i think u should be \[3x ^{2}-3x\]

Answer 18

yeah it should , keep going :)

Answer 19

\[u=3x ^{2}-3x\]\[=>du=6x-3 dx\]\[\div 3 => \frac{ 1 }{ 2 }du = 2x-1 dx\]
so
\[\frac{ 1 }{ 2 } \int\limits_{}^{} \frac{ du }{ \sqrt{u} } ? \]

Answer 20

sorry, mistake instead of 1/2 it is 1/3

Answer 21

yes u got it right its 1/3

Answer 22

\[\frac{ 1 }{ 3 } \int\limits_{}^{} \frac{ du }{ \sqrt{u} }\]
didnt know how to get its antiderivative :/

Answer 23

oh, wait

Answer 24

you dont know how to integrate \[\int\limits_{}^{}{du/\sqrt{u}}\]
?

Answer 25

\[=>\frac{ 1 }{ 3 } \int\limits_{}^{} \frac{ 1 }{ \sqrt{u} } du\]
\[=>\frac{ 1 }{ \sqrt{u} } = 2\sqrt{u} \] so
\[=>\frac{ 2\sqrt{u} }{ 3 } + c\]
\[=> \frac{ 2\sqrt{3x ^{2}-3x} }{ 3 } +c ?\]

not really sure about this

Answer 26

@zepdrix Hello :D

Answer 27

ok why dont you first rewrite \[1/\sqrt{u}=u^{-1/2}\]

Answer 28

ok, \[so (3x^{2}-3x)^{\frac{ -1 }{ 2 }}\]\[=>\frac{ 1 }{ 2 }(3x ^{2}-3x)^{\frac{ 1 }{ 2 }}(6x-3) + c ?\]

Answer 29

when u integrate \[u^{-1/2}\] you should get \[2*u^{1/2}\] do you see how ?

Answer 30

no, i followed this formulae \[\int\limits_{}^{} x ^{n} dx = \frac{ 1 }{ n+1 }x ^{n+1}+c\]

Answer 31

well yes if you followed that u should get that what i said since \[x^{-1/2+1}/(1/2)=2*x^{}\]

Answer 32

\[2*x^{1/2}\] mistake up there

Answer 33

this is really confusing :/

Answer 34

why is it confusing you just follow the formula , n=-1/2 and substitude for that in the fomula

Answer 35

ok, so \[2u ^{\frac{ 1 }{ 2 }}\]
\[=> 2(3x ^{2}-3x)^{\frac{ 1 }{ 2 }}\]

Answer 36

2 should be 3 right?

Answer 37

if u remember we had \[\frac{ 1 }{3 }\int\limits_{}^{}{1/\sqrt{u}}=\frac{ 2 }{ 3 }*u^{1/2}\]

Answer 38

can you rewrite the steps again if you dont mind, i am little but confused

Answer 39

\[\int\limits_{}^{}{(2x-1)/(3x^2-3x)^{1/2}}\]
we use substitution u=3x^2-3x so we have du=(6x-3)dx=3(2x-1)dx=>du/3=2x-1

Answer 40

so we have \[\frac{ 1 }{ 3 }\int\limits_{}^{}{du/\sqrt{u}}\]

Answer 41

then we have
\[\frac{ 1 }{ 3 }*\int\limits_{}^{}{u^{-1/2}}=\frac{ 1 }{ 3 }*\frac{ u^{(-1/2)+1} }{ (-1/2)+1 }=\frac{ 1 }{3 }*\frac{ u^{1/2} }{ 1/2 }=\frac{ 2 }{ 3 }*u^{1/2}+C\]

Answer 42

what happen to du in \[\frac{ du }{ \sqrt{u} }\]

Answer 43

oh i missed it , should be there but that doesnt change the solution after its still that way

Answer 44

and now i totaly have to go , good luck :) if u need more hep post other question someone else will help you

Answer 45

thank you very much

Answer 46

no problem :)