Question

calculus trig substitution

Answer 1

\[\int\limits \frac{ x^5 }{ \ \sqrt{x^2+2} }dx\]

Answer 2

would x = tanθ ?

Answer 3

\[\sqrt{a^2+x^2} \]
substitution: \[x = a \tan \theta \]
\[dx = a \sec^2\theta d \theta\]

Answer 4

identity : \[1+\tan ^{2}\theta = \sec^2\theta \]

Answer 5

so you agree a^2 = 2 here?

Answer 6

thats what i have in my notes

Answer 7

a = sqrt (2)

Answer 8

yes
so what would you substitute?

Answer 9

\(x= \sqrt 2 \tan t\)
right?

Answer 10

yes

Answer 11

\[dx = \sqrt{2} \sec^2 \theta d \theta \]

Answer 12

Looks good

Answer 13

okay, what do i do next?

Answer 14

;

Answer 15

Since you have the x and dx, we should plug it into the original equation

Answer 16

plug whatever you know in the original integral

Answer 17

please try with this substitution:
\[\huge x = \sqrt 2 \sinh \theta \]

Answer 18

why sin?

Answer 19

since we have these steps:
\[{x^2} + 2 = 2{\left( {\sinh \theta } \right)^2} + 2 = 2\left( {1 + {{\left( {\sinh \theta } \right)}^2}} \right) = 2{\left( {\cosh \theta } \right)^2}\]

Answer 20

what's the h?

Answer 21

after that variable change, we get:
\[\Large \int {\frac{{{x^5}}}{{\sqrt {{x^2} + 2} }}dx = } {2^{5/2}}\int {{{\left( {\sinh \theta } \right)}^5}d\theta } \]

Answer 22

it means for hyperbolic function

Answer 23

can i try it the first way? because that's how my teacher showed us

Answer 24

ok!

Answer 25

I think that we can make this substitution:
\[x = 2\tan \theta \]

Answer 26

oops.. I meant:
\[\huge x = \sqrt 2 \tan \theta \]

Answer 27

\[\int\limits \frac{ (\sqrt{2}\tan \theta)^5(\sqrt{2}\sec^2\theta) }{ \sqrt{(\sqrt{2}\tan \theta )^2+2} }d \theta \]

Answer 28

yes! I got this:
\[\Large {2^{5/2}}\int {\frac{{{{\left( {\tan \theta } \right)}^5}}}{{\cos \theta }}d\theta } \]

Answer 29

how would i get that?

Answer 30

we have the subsequent steps:
\[\Large \begin{gathered}
{2^{5/2}}\int {\frac{{{{\left( {\tan \theta } \right)}^5}}}{{\cos \theta }}d\theta } = \hfill \\
\hfill \\
= {2^{5/2}}\int {\frac{{{{\left( {\sin \theta } \right)}^5}}}{{{{\left( {\cos \theta } \right)}^6}}}d\theta } = {2^{5/2}}\int {\frac{{{{\left( {\sin \theta } \right)}^4}}}{{{{\left( {\cos \theta } \right)}^6}}}\sin \theta d\theta } = \hfill \\
\hfill \\
= {2^{5/2}}\int {\frac{{{{\left( {1 - {{\left( {\cos \theta } \right)}^2}} \right)}^2}}}{{{{\left( {\cos \theta } \right)}^6}}}\sin \theta d\theta } \hfill \\
\end{gathered} \]

Answer 31

ok, how can i simplify what i got?

Answer 32

we can make this varaible change:
\[\Large u = \cos \theta \]

Answer 33

so, we get:
\[\Large \begin{gathered}
{2^{5/2}}\int {\frac{{{{\left( {\tan \theta } \right)}^5}}}{{\cos \theta }}d\theta } = \hfill \\
\hfill \\
= {2^{5/2}}\int {\frac{{{{\left( {\sin \theta } \right)}^5}}}{{{{\left( {\cos \theta } \right)}^6}}}d\theta } = {2^{5/2}}\int {\frac{{{{\left( {\sin \theta } \right)}^4}}}{{{{\left( {\cos \theta } \right)}^6}}}\sin \theta d\theta } = \hfill \\
\hfill \\
= {2^{5/2}}\int {\frac{{{{\left( {1 - {{\left( {\cos \theta } \right)}^2}} \right)}^2}}}{{{{\left( {\cos \theta } \right)}^6}}}\sin \theta d\theta } = \hfill \\
\hfill \\
= - {2^{5/2}}\int {\frac{{{{\left( {1 - {u^2}} \right)}^2}}}{{{u^6}}}du} \hfill \\
\end{gathered} \]

Answer 34

and now it is easy to solve such integral

Answer 35

@darkigloo u still doing this??

Answer 36

yes, how do i simplify what i got? what happened to the radical 2?

Answer 37

@rishavraj

Answer 38

u going with wht michelelaino mentioned??

Answer 39

i don't really get how he simplified what i got
\[\int\limits \frac{ (\sqrt2\tan \theta)^5(\sqrt2\sec^2\theta )}{ \sqrt{(\sqrt{2}\tan \theta)^2+2} }d \theta\]

Answer 40

see I am going this way
\[\int\limits \frac{ 4\sqrt{2}\tan^5 \theta }{ \sqrt{2\tan^2 \theta + 2} }\]
\[\int\limits \frac{ 4\sqrt2\tan^3 \theta(\sec^2 \theta - 1) }{ \sqrt{2\tan^2 \theta + 2} }\]

what say???@darkigloo

Answer 41

how did you get that?
for what i got, i only plugged in x. is that right?

Answer 42

see plug \[x = \sqrt 2 \tan \theta\]
u will get tht :))

Answer 43

and i plug in \[ dx = \sqrt{2} \sec^2\theta d \theta \] too ?

Answer 44

ooo sucks my bad i didnt make tht change ...sorrry ..... thts correct

Answer 45

@darkigloo gimme few mins lemme solve it and then i would let u know....
and also are u supposd to use trig substitution??

Answer 46

yes

Answer 47

ok hold on :))

Answer 48

ok thanks :)

Answer 49

so it would be \[\int\limits \frac{ 4\sqrt2 \tan^5 \theta \sqrt2 \sec^2 \theta }{ \sqrt{2 \tan^2 \theta + 2} }d theta\]

\[\int\limits \frac{ 4\sqrt2 \tan^5 \theta \sqrt2 \sec^2 \theta }{ \sqrt{2 (\tan^2 \theta + 1)} } d \theta\]

Answer 50

then
\[\int\limits \frac{ 4\sqrt2 \tan^5 \theta \sqrt2 \sec^2 \theta }{ \sqrt{2}\sec \theta } d \theta\]

Answer 51

@darkigloo ///????

Answer 52

where did the 4 come from?

Answer 53

lemme know if u find any error....
\[(\sqrt2 \tan \theta)^5 = 4\sqrt2 \tan^5 \theta ~~~ ?????\]

Answer 54

hmmmm ??? @darkigloo

Answer 55

oh i see )

Answer 56

@darkigloo done :)))) :P

Answer 57

....

Answer 58

how does the denominator \[\sqrt{(\sqrt{2}\tan \theta)^2+2}\] become
\[\sqrt{2\tan^2\theta+1}\]

Answer 59

its not tht
\[\sqrt{(\sqrt2\tan \theta)^2 + 2} = \sqrt{2\tan^2 \theta + 2} = \sqrt{2(\tan^2 \theta + 1)} = \sqrt{2 \sec^2 \theta} = \sqrt2 \sec \theta\]

Answer 60

so it would be..(I am gonna plug the denominator as above)
\[\int\limits \frac{ 4\sqrt2 \tan^5 \theta \sqrt2 \sec^2 \theta }{ \sqrt{2}\sec \theta } d \theta\]

Answer 61

@darkigloo now u can proceed???

Answer 62

\[\int\limits \frac{ 4\tan^5\theta \sqrt{2}\sec^2\theta }{ \sqrt{2}\sec \theta } d \theta = \int\limits 4\tan^5\theta \sqrt{2}\sec \theta d \theta \]

Answer 63

good work :))) proceed ...and yeah u reply really late -_- no offense
or write it like
\[\int\limits \frac{ 4\sqrt2 \tan^5 \theta ~ \sec \theta }{ \sqrt{2}\sec \theta } d \theta\]

Answer 64

haha sorry i'm kinda busy. im about to leave.
why do you get \[\frac{ \sec \theta }{ \sqrt{2}\sec \theta }\]

Answer 65

oooops my bad...i forgot to delete tht....... sorry it wasnt suppsed to be ther

Answer 66

the denominator isnt supposed to be there?

Answer 67

it would be
\[\int\limits \frac{ 4\sqrt2 \tan^5 \theta \sqrt2 \sec^2 \theta }{ \sqrt2 \sec \theta } d \theta\]

\[\int\limits { 4\sqrt2 \tan^5 \theta ~ \sec \theta }~~ d \theta\]

Answer 68

there u go ...now u can split it like
\[\int\limits 4\sqrt2 \tan^3 \theta \tan^2 \theta \sec \theta ~d \theta \]

\[\int\limits 4\sqrt2 \tan^3 \theta~ (\sec^2 \theta - 1) \sec \theta ~d \theta \]

just rearrangement
\[\int\limits 4\sqrt2 \tan^3 \theta \sec \theta~ (\sec^2 \theta - 1) ~d \theta \]

Answer 69

now
\[\int\limits (4\sqrt2 \tan^3 \theta \sec^3 \theta - 4\sqrt2 \tan^3 \theta \sec \theta) ~~d \theta\]
so

\[\int\limits 4\sqrt2 \tan^3 \theta \sec^3 \theta~~d \theta ~~- ~~\int\limits 4\sqrt2 \tan^3 \theta \sec \theta ~~d \theta\]

Answer 70

@darkigloo ?????

Answer 71

ok for when u get here..... @darkigloo
\[4 \sqrt2 \int\limits \tan^2 \theta \tan \theta \sec^3 \theta ~d \theta ~-~ 4 \sqrt2 \int\limits \tan^2 \theta \tan \theta \sec \theta ~ d \theta\]

Answer 72

solve both the integrals :)) lemme know if u got any further doubt :))) @darkigloo

Answer 73

@darkigloo u there???

Answer 74

@rishavraj im here now :)

Answer 75

will u be able to solve it now?

Answer 76

i just got online, im trying now

Answer 77

ok lemme know if u get any doubt :))) @darkigloo

Answer 78

im having trouble, will i have to choose a u now?

Answer 79

see now u got 2 integrals...solve them separately ......and yeah u need a 'u'

Answer 80

hold on I will solve the first integral .....see its
\[4\sqrt2 \int\limits \tan^3 \theta \sec^3 \theta ~d \theta\]
\[4\sqrt2 \int\limits \tan^2 \theta \tan \theta \sec^3 \theta ~d \theta\]
\[4\sqrt2 \int\limits (\sec^2 \theta - 1)\tan \theta \sec^3 \theta ~d \theta\]

now plug


so it would be now
\[4\sqrt2\int\limits u^2(u^2 - 1)~du\]

Answer 81

what is u= ?

Answer 82

oops my bad
\[\sec \theta = u\]

Answer 83

as \[\sec \theta ~=~u\]
so
\[\sec \theta ~\tan \theta~ d \theta = du\]

Answer 84

@darkigloo i am only solving the first integral....not the second one among the two :)))

Answer 85

ok i think i get it. i'll try the second one.

Answer 86

good then :)))

Answer 87

\[4\sqrt{2} \int\limits \tan^2\theta \tan \theta \sec \theta d \theta \]
\[= 4\sqrt{2} \int\limits \ (sec^2 - 1) \tan \theta \sec \theta d \theta \]
\[4 \sqrt 2 \int\limits(u^2 -1)du\]

Answer 88

@darkigloo good work ;)))

Answer 89

yay! now ill try to continue.

Answer 90

go on :))

Answer 91

for the first one, im pretty sure im wrong. :(
\[64\sqrt2 \int\limits u^2 (u^2-1) = 4\sqrt2 (\frac{ u^3 }{ 3 }(\frac{ u^3 }{ 3 }-\frac{ u^2 }{ 2 })+c\]

Answer 92

not 64, i meant 4

Answer 93

for first part it would be
\[4\sqrt2 (\frac{ u^5 }{ 5 } ~ - ~\frac{ u^3 }{ 3 }) \]
then as u = sec theta so
\[4\sqrt2 (\frac{ \sec^5 \theta }{ 5 } ~-~\frac{ \sec^3 \theta}{ 3 })\]

Answer 94

see thts simple......
\[4\sqrt2 \int\limits u^2(u^2 - 1) ~du ~=~ 4\sqrt2 \int\limits u^4 - u^2 ~du\]

Answer 95

ahh i see :)

Answer 96

for the second one :
\[4 \sqrt 2 \int\limits (u^2 -1)du = 4\sqrt{2}\frac{ u^3 }{ 3 }-u +c \]
\[= 4 \sqrt2 \frac{ \sec^3 \theta }{ 3}-\sec \theta +c\]