Why does the LIATE rule not work here? (1 Viewer)

Run hard@thehsc · May 29, 2022

I rely on the LIATE and realised it does not work here? Can someone clarify....

ExtremelyBoredUser · May 29, 2022

Run hard@thehsc said:
View attachment 35690
I rely on the LIATE and realised it does not work here? Can someone clarify....

I'm not sure how the LIATE rule doesn't work? You can choose to integrate exponential first then trig or alternately

$u = e^{2x} \Rightarrow du = 2e^{2x} dx , dv = \cos{(x)} \, dx \Rightarrow v = \sin{(x)}$

$I = \left(\left[e^{2x}\sin{(x)}\right\]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2e^{2x}\sin{(x)} \,dx \right)$
$I = e^{\pi} - \int_{0}^{\frac{\pi}{2}} 2e^{2x}\sin{(x)} \, dx$

$u = 2e^{2x} \Rightarrow du = 4e^{2x} dx , dv = \sin{(x)} \, dx \Rightarrow v = -\cos{(x)}$

$I = e^{\pi} - \left( \left[-2e^{2x}\cos{(x)}\right]_{0}^{\frac{\pi}{2}} + 4 \int_{0}^{\frac{\pi}{2}} e^{2x}\cos{(x)} \,dx \right)$
$I = e^{\pi} - ( - 0 - (-2) + 4I)$
$I = e^{\pi} - 2 - 4I$
$5I = e^{\pi} - 2$

$I = \frac{e^{\pi} - 2}{5}$

I don't really go off LIATE since integrating the exp. here feels a bit messy dealing with fractions and I'd rather just deal with whole numbers. You'll get the same thing. Personally I'd say use LIATE as a guide but don't be bounded by it, use your own intuition to foresee which one's going to be more convenient to go through with IBP, it helps reduce algebraic errors.

mmmmmmmmaaaaaaa · May 29, 2022

I don't know what you've chose to integrate and differentiate, however,

as @ExtremelyBoredUser stated, you should get your answer by letting u = e^(2x) and v'(dv) = cosx

yanujw · May 29, 2022

LIATE is the order for the functions that are best to integrate. cosx is a Trig function therefore is better to integrate, while e^2x is Exponential so better to differentiate.

ekjchale#1 · May 29, 2022

yanujw said:
LIATE is the order for the functions that are best to integrate. cosx is a Trig function therefore is better to integrate, while e^2x is Exponential so better to differentiate.

is it possible for u to recall the full form of liate as ive forgotten

yanujw · May 29, 2022

ekjchale#1 said:
is it possible for u to recall the full form of liate as ive forgotten

Personally I remember the order DETAIL which is a similar concept

Differentiate to the following priority;
Exponentials
Trig functions
Algebra (polynomials)
Inverse Trig functions
Logarithms

ekjchale#1 · May 29, 2022

yanujw said:
Personally I remember the order DETAIL which is a similar concept

Differentiate to the following priority;
Exponentials
Trig functions
Algebra (polynomials)
Inverse Trig functions
Logarithms

aye thx a lot bro

Run hard@thehsc · May 30, 2022

ExtremelyBoredUser said:
I'm not sure how the LIATE rule doesn't work? You can choose to integrate exponential first then trig or alternately

$u = e^{2x} \Rightarrow du = 2e^{2x} dx , dv = \cos{(x)} \, dx \Rightarrow v = \sin{(x)}$

$I = \left(\left[e^{2x}\sin{(x)}\right\]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2e^{2x}\sin{(x)} \,dx \right)$
$I = e^{\pi} - \int_{0}^{\frac{\pi}{2}} 2e^{2x}\sin{(x)} \, dx$

$u = 2e^{2x} \Rightarrow du = 4e^{2x} dx , dv = \sin{(x)} \, dx \Rightarrow v = -\cos{(x)}$

$I = e^{\pi} - \left( \left[-2e^{2x}\cos{(x)}\right]_{0}^{\frac{\pi}{2}} + 4 \int_{0}^{\frac{\pi}{2}} e^{2x}\cos{(x)} \,dx \right)$
$I = e^{\pi} - ( - 0 - (-2) + 4I)$
$I = e^{\pi} - 2 - 4I$
$5I = e^{\pi} - 2$

$I = \frac{e^{\pi} - 2}{5}$

I don't really go off LIATE since integrating the exp. here feels a bit messy dealing with fractions and I'd rather just deal with whole numbers. You'll get the same thing. Personally I'd say use LIATE as a guide but don't be bounded by it, use your own intuition to foresee which one's going to be more convenient to go through with IBP, it helps reduce algebraic errors.

ye fair it does work - the worked sols seemed rigged as they let u = cos instead.......

Why does the LIATE rule not work here? (1 Viewer)

Run hard@thehsc

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ExtremelyBoredUser

Bored Uni Student

mmmmmmmmaaaaaaa

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yanujw

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ekjchale#1

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yanujw

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ekjchale#1

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Run hard@thehsc

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