JEE Exam > JEE Videos > Mathematics (Maths) Class 12 > Examples : Integration by Parts 1
Examples : Integration by Parts 1 Video Lecture | Mathematics (Maths) Class 12 - JEE
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FAQs on Examples : Integration by Parts 1 Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. What is integration by parts and how does it work? |  |
Ans. Integration by parts is a technique used in calculus to evaluate the integral of a product of two functions. It is based on the product rule for differentiation. The formula for integration by parts is ∫ u dv = uv - ∫ v du, where u and v are differentiable functions. By selecting appropriate u and dv, we can simplify the integral by integrating v and differentiating u.
| 2. When should I use integration by parts? | |
Ans. Integration by parts is typically used when we have a product of two functions in the integrand. It is particularly helpful when one of the functions becomes simpler or easier to integrate after differentiating it, while the other function becomes simpler or easier to differentiate after integrating it. It is also useful when the integral involves trigonometric functions, logarithmic functions, or exponential functions.
| 3. Can you provide an example of integration by parts? | |
Ans. Certainly! Let's consider the integral ∫ x sin(x) dx. To solve this, we can choose u = x and dv = sin(x) dx. Then, du = dx and v = -cos(x). Applying the integration by parts formula, we have:
∫ x sin(x) dx = -x cos(x) - ∫ (-cos(x)) dx = -x cos(x) + sin(x) + C,
where C is the constant of integration.
| 4. Are there any specific rules or guidelines to follow when applying integration by parts? | |
Ans. Yes, there are guidelines to follow when applying integration by parts:
1. Choose u and dv: Select u as a function that becomes simpler after differentiating, and dv as a function that becomes simpler after integrating.
2. Calculate du and v: Differentiate u to find du and integrate dv to find v.
3. Apply the integration by parts formula: Using the formula ∫ u dv = uv - ∫ v du, substitute the values of u, dv, du, and v into the formula.
4. Simplify and solve: Evaluate the integral on the right-hand side of the formula, simplifying as much as possible, and add the constant of integration if necessary.
| 5. Can integration by parts be used to solve definite integrals? | |
Ans. Yes, integration by parts can be used to solve definite integrals as well. After applying integration by parts to obtain a simplified expression, we can substitute the limits of integration and evaluate the definite integral. Just make sure to keep track of any additional terms that may arise from the integration by parts process, such as boundary terms, and include them in the final evaluation.
Text Transcript from Video
hello friends this video on integrals
part 31 is brought you by exam fear calm
no words away from exam if we're
watching this video please make sure
that you have watched part 1 the power
30 this is a special type of integration
by paths if you have this function of
this form e to the power x FX plus FS x
DX that will be coming into power ik x
FX i prove how to do this let me break
this i can bind this has nothing but e
to the power x FX DX class e to the
power X M des amis so as I discussed in
integration by parts the choice of first
function is very critical so let's have
a memory team Pilate right inverse
longer to make algebraic digna metric
exponential exponential consider the
last option right so I'll take this guy
is second and this guy is first
functional why because exponential at
the last possible last option for us
see if I take this guy's first function
my equation is f X integration of e^x DX
right minus f dash X integration of e^x
DX the whole thing integrated plus
integration of e to the power a X F dash
X yes right so this becomes F X
integration of e to the power X DX is e
to the power X itself minus f dash X T
to the power X DX integration is e to
the power X itself and this guy
integrated DX plus e to the power X F
dash X you see both are same this guy is
the essence so cut those legs so what I
am left to this FX e to the power X plus
obviously one constant lad here because
the moment I'll differentiate this guy
or integrate this guy I will get a
constant because out here
here so and that is what my Irish is
this
so I prove that this type of spatial
integration where they have been
debarred XFX + f nest x then it is
nothing but into particles now let's
solve solution X square e to the power X
D is defined let's have an ability
violate inverse is not there logarithmic
not there
algebraic yes I have it so this day my
first in this back in a second
so this guy will become first X square
integration of e to the power X DX minus
D by DX of X square integration of e to
the power X DX the whole thing again we
have to integrate with DX that is
nothing but X square e to the power X
right minus X square integration is 2x e
to the power X DX sorry now again if you
see this is also same multiplication of
two function integral same form formula
applied do have inverse know and logger
to make no algebraic yes I think this
guy is punched is the x ii + c obviously
will come here now so this there will
become X square e to the power X - now I
think in this fashion 2 X and the
constant integration of e to the power X
DX
- derivative of 2 X 3 by DX of 2 X into
integration of e to the power X DX the
whole thing is again integrated that's
the formula plus C this becomes X square
e to the power X minus 2 mix e to the
power X DX into bar X itself right minus
D by D's of tomato comes to e to the
power X DX and the whole thing is
integrated this guy is nothing but X
square e to the power X minus 2 XE to
the power X
- 2 e to the power X DX sorry DX is gone
huh the collection because this guy is e
to the power x + mountain so you take
either by the XCOM and this becomes e to
the bar X into X square minus 2 X 2 plus
in this mask very simple application of
integration by parts and the reason why
we applied again twice is because we
know that this was X square this become
X and I knew next time
this day will become constant correctly
let's take for example X log it it's
again this formula or memory tip Pilate
do you have any inverse no logarithmic
yes I have so this guy becomes 1 this
Geist I think this guy is dot gives
integration formula is FX GX integration
is nothing but FX integration of GX
minus integration of F dash X GX DX
right this is the same thing will apply
here so log X into integration of X
right DX minus dy by DX sorry
integration of D by DX of log X into
integration of X DX
correct the whole thing is integrated
with respect to X so this becomes log X
into into X square by 2 minus D by D's
of log X 1 by X and this integration of
X DX is X square by 2 right and the
whole thing is integrated with respect
not
the X plus constant so this becomes X
square by 2 log X minus the cancels 1 by
2 X DX is again X square by 2 square
plus constant we solve this further is
nothing but X square by 2 log X minus X
square by 4 and that is matter where is
it but just to remember this formula
aisle 8 you know which one is first
which means second in this case not X
was first X was second so I love X log X
into indication of X DX minus D by D's
of log X integration of FDX the whole
thing again integrated is respect to X
and then solve it to get the answer
let's find problem x Square log is this
was X log X this X square bollocks
almost see here also will form have the
formula
aisle 8 do we have inverse no
logarithmic yes this guy became first
second so insane integration of log X X
square DX this is nothing but log X
integration of X square DX minus D by DX
of X into integration of X square DX
whatever you get the whole thing is
integrated with respect to DX so this
becomes log X into this is X square this
becomes X cube by 3 minus log X
Direction is 1 by X into this guy
becomes X cube by 3 DX right this is
nothing but X but not canceled so what
you get here is unlike properly so then
you can change you can expose xq x cube
by X in X square so X square by 2 DX
this is nothing but log X or X cube by 3
X cube by 3 log X minus integration of 1
by 2 X square
this is nothing but X cubed 3 log X
minus this becomes X cube by 3 3 3 6 + 1
was that is my answer correct so this is
X cube by 3 this is 3 only & 3 & 3 is 9
and as minus X cube by 9 so very simple
not a great Rock resize it the critical
part in this kind of cushion is you
should remember
I'll eat find the first function and
just use the formula normal formula to
solve the ocean let's add this some more
example X sine inverse X again let us
have this formula aisle 8 do we have
inverse yes we have this guys first this
guy's sitting so I have to find sine
inverse X DX
okay so this there will be nothing but
sine inverse takes into integral of x DX
minus D by DX of sine inverse X integral
of xdx the whole thing whatever you get
is the integral of this right sine
inverse X you'll write constant X DX
becomes X square by 2 minus integration
of differentiation of sine personal
thing but root 1 minus 1 by X square and
this is X DX will become X square by D
the whole thing we have to integrate
sorry now this kind of potion we have
solved correct we have X square by root
1 minus X square so what we can do is we
can say 1 minus X square and we can plus
-1 we can so if you see this way this is
nothing but sly inverse X X square by
2-a right here oniy right plus is minus
X square right 1 minus X square plus 1
minus 1 so 1 minus 1 is 0 and minus X
square will remain here so but this what
happens is 1 by 2 anyway a common
yes so what I have done here I rewrote
in this fashion because this I can write
in my term which I like because see one
this kind of formal I know you if it is
one I know how to find this so I won I
did worse this was X square so I say 1
minus X square minus X square 1 minus 1
is 18 is 0 but this I can factorize
further and this formula I know how to
find integration right so this is
nothing but X square so I inverse X plus
1 by 2 integral of you see this become
nothing but root of 1 minus X square DX
and this is nothing but minus 1 by 2
integral of 1 by 1 minus X square DX
correct now we know how to find the
integration of this integration of this
will do that it's nothing but X square
sine inverse X plus 1 by 2 this is what
integration of 1 minus X square DX is
nothing but and right here 1 by 2 or X
by 2 root 1 minus 6 square correct plus
1 by 2 sine inverse X minus this guy
becomes this is nothing but sine inverse
X plus some constant C this is all this
further this is nothing but X square by
2 sine inverse X in this you see this is
1 by 4 sine inverse X minus 1 by 2 so it
is plus 1 by 4 sine inverse X right and
it is plus X by 4 root 1 minus X square
plus some constant thank you
visit exam Fiat
come to watch free educational videos
try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thanks and screen
part 31 is brought you by exam fear calm
no words away from exam if we're
watching this video please make sure
that you have watched part 1 the power
30 this is a special type of integration
by paths if you have this function of
this form e to the power x FX plus FS x
DX that will be coming into power ik x
FX i prove how to do this let me break
this i can bind this has nothing but e
to the power x FX DX class e to the
power X M des amis so as I discussed in
integration by parts the choice of first
function is very critical so let's have
a memory team Pilate right inverse
longer to make algebraic digna metric
exponential exponential consider the
last option right so I'll take this guy
is second and this guy is first
functional why because exponential at
the last possible last option for us
see if I take this guy's first function
my equation is f X integration of e^x DX
right minus f dash X integration of e^x
DX the whole thing integrated plus
integration of e to the power a X F dash
X yes right so this becomes F X
integration of e to the power X DX is e
to the power X itself minus f dash X T
to the power X DX integration is e to
the power X itself and this guy
integrated DX plus e to the power X F
dash X you see both are same this guy is
the essence so cut those legs so what I
am left to this FX e to the power X plus
obviously one constant lad here because
the moment I'll differentiate this guy
or integrate this guy I will get a
constant because out here
here so and that is what my Irish is
this
so I prove that this type of spatial
integration where they have been
debarred XFX + f nest x then it is
nothing but into particles now let's
solve solution X square e to the power X
D is defined let's have an ability
violate inverse is not there logarithmic
not there
algebraic yes I have it so this day my
first in this back in a second
so this guy will become first X square
integration of e to the power X DX minus
D by DX of X square integration of e to
the power X DX the whole thing again we
have to integrate with DX that is
nothing but X square e to the power X
right minus X square integration is 2x e
to the power X DX sorry now again if you
see this is also same multiplication of
two function integral same form formula
applied do have inverse know and logger
to make no algebraic yes I think this
guy is punched is the x ii + c obviously
will come here now so this there will
become X square e to the power X - now I
think in this fashion 2 X and the
constant integration of e to the power X
DX
- derivative of 2 X 3 by DX of 2 X into
integration of e to the power X DX the
whole thing is again integrated that's
the formula plus C this becomes X square
e to the power X minus 2 mix e to the
power X DX into bar X itself right minus
D by D's of tomato comes to e to the
power X DX and the whole thing is
integrated this guy is nothing but X
square e to the power X minus 2 XE to
the power X
- 2 e to the power X DX sorry DX is gone
huh the collection because this guy is e
to the power x + mountain so you take
either by the XCOM and this becomes e to
the bar X into X square minus 2 X 2 plus
in this mask very simple application of
integration by parts and the reason why
we applied again twice is because we
know that this was X square this become
X and I knew next time
this day will become constant correctly
let's take for example X log it it's
again this formula or memory tip Pilate
do you have any inverse no logarithmic
yes I have so this guy becomes 1 this
Geist I think this guy is dot gives
integration formula is FX GX integration
is nothing but FX integration of GX
minus integration of F dash X GX DX
right this is the same thing will apply
here so log X into integration of X
right DX minus dy by DX sorry
integration of D by DX of log X into
integration of X DX
correct the whole thing is integrated
with respect to X so this becomes log X
into into X square by 2 minus D by D's
of log X 1 by X and this integration of
X DX is X square by 2 right and the
whole thing is integrated with respect
not
the X plus constant so this becomes X
square by 2 log X minus the cancels 1 by
2 X DX is again X square by 2 square
plus constant we solve this further is
nothing but X square by 2 log X minus X
square by 4 and that is matter where is
it but just to remember this formula
aisle 8 you know which one is first
which means second in this case not X
was first X was second so I love X log X
into indication of X DX minus D by D's
of log X integration of FDX the whole
thing again integrated is respect to X
and then solve it to get the answer
let's find problem x Square log is this
was X log X this X square bollocks
almost see here also will form have the
formula
aisle 8 do we have inverse no
logarithmic yes this guy became first
second so insane integration of log X X
square DX this is nothing but log X
integration of X square DX minus D by DX
of X into integration of X square DX
whatever you get the whole thing is
integrated with respect to DX so this
becomes log X into this is X square this
becomes X cube by 3 minus log X
Direction is 1 by X into this guy
becomes X cube by 3 DX right this is
nothing but X but not canceled so what
you get here is unlike properly so then
you can change you can expose xq x cube
by X in X square so X square by 2 DX
this is nothing but log X or X cube by 3
X cube by 3 log X minus integration of 1
by 2 X square
this is nothing but X cubed 3 log X
minus this becomes X cube by 3 3 3 6 + 1
was that is my answer correct so this is
X cube by 3 this is 3 only & 3 & 3 is 9
and as minus X cube by 9 so very simple
not a great Rock resize it the critical
part in this kind of cushion is you
should remember
I'll eat find the first function and
just use the formula normal formula to
solve the ocean let's add this some more
example X sine inverse X again let us
have this formula aisle 8 do we have
inverse yes we have this guys first this
guy's sitting so I have to find sine
inverse X DX
okay so this there will be nothing but
sine inverse takes into integral of x DX
minus D by DX of sine inverse X integral
of xdx the whole thing whatever you get
is the integral of this right sine
inverse X you'll write constant X DX
becomes X square by 2 minus integration
of differentiation of sine personal
thing but root 1 minus 1 by X square and
this is X DX will become X square by D
the whole thing we have to integrate
sorry now this kind of potion we have
solved correct we have X square by root
1 minus X square so what we can do is we
can say 1 minus X square and we can plus
-1 we can so if you see this way this is
nothing but sly inverse X X square by
2-a right here oniy right plus is minus
X square right 1 minus X square plus 1
minus 1 so 1 minus 1 is 0 and minus X
square will remain here so but this what
happens is 1 by 2 anyway a common
yes so what I have done here I rewrote
in this fashion because this I can write
in my term which I like because see one
this kind of formal I know you if it is
one I know how to find this so I won I
did worse this was X square so I say 1
minus X square minus X square 1 minus 1
is 18 is 0 but this I can factorize
further and this formula I know how to
find integration right so this is
nothing but X square so I inverse X plus
1 by 2 integral of you see this become
nothing but root of 1 minus X square DX
and this is nothing but minus 1 by 2
integral of 1 by 1 minus X square DX
correct now we know how to find the
integration of this integration of this
will do that it's nothing but X square
sine inverse X plus 1 by 2 this is what
integration of 1 minus X square DX is
nothing but and right here 1 by 2 or X
by 2 root 1 minus 6 square correct plus
1 by 2 sine inverse X minus this guy
becomes this is nothing but sine inverse
X plus some constant C this is all this
further this is nothing but X square by
2 sine inverse X in this you see this is
1 by 4 sine inverse X minus 1 by 2 so it
is plus 1 by 4 sine inverse X right and
it is plus X by 4 root 1 minus X square
plus some constant thank you
visit exam Fiat
come to watch free educational videos
try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thanks and screen
About this Video
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Nov 21, 2024 Last updated |
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