Integration by Parts: Z U Dv = Uv Z V Du Proof: This Is Proved Integrating The

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Integration by Parts: udv = uv vdu Z Z Proof: This is proved integrating the product rule. The product rule states (fg)0 = f 0g + fg 0.

Integrating: (fg)0 dx = f 0g + fg 0 dx. Z Z Simplifying: fg = f 0gdx+ fg 0 dx. Z Z Rearranging: fg 0 dx = fg f 0gdx. Z Z Plug in u = f (x)andv = g(x)sothatdu = f 0(x)dx and dv = g 0(x)dx to get udv = uv vdu. Z Z Steps to use Integration by Parts: First choose a suitable u. Guidelines - Use the following table to choose u starting at the top (works 95% of the time). 1 inverse trig function 2 logarithm 3 polynomial, root function or expression with powers of x 4 trigonometric function After choosing u,letdv be the leftover part in the integral (including dx). Fill in the boxes: u = dv = ··· ··· derivative 8 9 integral > du = v = > < ··· ··· = Plug into the formula! :> ;> Math 267 (University of Calgary) Fall 2015 1 / 9 The LIATE rule

Alternate guidelines to choose u for integration by parts was proposed by H. Kasube. (See the article: Kasube, Herbert E. A Technique for Integration by Parts.PublishedinThe American Mathematical Monthly Volume 90 (3), 1983, pages 210–211.) It uses the acronym LIATE and which ever function comes ﬁrst in the list should be chosen for u: 1 L -Logarithmicfunctions

F ln x,logb x,ln2x,etc. 2 I -Inversetrigonometricfunctions 1 1 1 F sin x,sec x,tan (2x), etc. 3 A -Algebraicfunctions(polynomials) 3 n F x +1,x ,5x +7,etc. 4 T -Trigonometricfunctions F sin x,sin(7x), cos x,etc. 5 E -Exponentialfunctions x 7x x F e , e ,19 ,etc. Notice that if you use this rule backwards to choose dv,thenyoutakedv to be whichever function comes last in the list: functions lower on the list have easier antiderivatives than the functions above them. This backwards rule of picking the dv is sometimes written as DETAIL where D stands for dv.

Math 267 (University of Calgary) Fall 2015 2 / 9 Example

Evaluate the following integral: xex dx. Z Solution: We use integration by parts: u dv = u v v du. Z Z We have no inverse trig functions or logarithms, thus we choose u to be x and dv the leftover part: u = x dv = ex dx

Observe that the integral u dv is exactly the integral we are trying to compute: x ex dx. Z Z The left side, we di↵erentiate. The right side, we integrate. u = x dv = ex dx derivative 8 9 integral <> du = dx v = ex dx = ex =>

When integrating, we can choose:> ANY anti-derivative! It is commonR to choose;>C =0. Now we plug into the formula, it helps to use arrows to remember it: u = x dv = ex dx derivative integral 8 & 9 <> du = dx v = ex =>

:> ;> x ex dx = x ex ex dx Z Z = xex ex + C Math 267 (University of Calgary) Fall 2015 3 / 9 Integrating the Logarithm Example

Evaluate the integral of the natural logarithm: ln xdx. Z Solution: We use integration by parts: u dv = u v v du. Z Z Using the guidelines, we choose u to be ln x and dv the leftover part: u =lnx dv = dx

Observe that the integral u dv is exactly the integral we are trying to compute: ln x dx. Z Z The left side, we di↵erentiate. The right side, we integrate. u =lnx dv = dx

derivative 8 & 9 integral > 1 > <> du = dx v = 1 dx = x => x > R > Now we plug into the formula::> ;> 1 ln xdx =lnx x x dx x Z Z = x ln x 1 dx Z = x ln x x + C Math 267 (University of Calgary) Fall 2015 4 / 9 Integrating an inverse trig function Example

1 Evaluate the integral of arcsine: sin xdx. Z Solution: We use integration by parts: u dv = u v v du. Z Z 1 Using the guidelines, we choose u to be sin x and dv the leftover part:

1 u =sin x dv = dx

The left side, we di↵erentiate. The right side, we integrate.

1 u =sin x dv = dx 8 9 derivative > & > integral > 1 > <> du = dx v = x => p1 x2 > > > > Now we plug into the formula::> ;>

1 1 1 sin xdx =sin x x x dx p1 x2 Z Z 1 x = x sin x dx p1 x2 Z =?

Math 267 (University of Calgary) Fall 2015 5 / 9 Solution (CONTINUED):

1 1 x sin xdx = x sin x dx p1 x2 Z Z =? x We now use substitution rule on the remaining integral: dx. p1 x2 Z Letting u be an inside function, we have: u =1 x2 Taking the derivative gives: du du = 2xdx dx = ! 2x Using this substitution we get: x x du 1 1 dx = = du p1 x2 pu ( 2x) 2 pu Z Z Z We rewrite the exponent and use the power rule for integration:

x 1 1/2 dx = u du p1 x2 2 Z Z 1 u1/2 = + C 2 1/2 = pu + C = 1 x2 + C The ﬁnal answer is then: p 1 1 sin xdx= x sin x + 1 x2 + C. Z p Math 267 (University of Calgary) Fall 2015 6 / 9 Example

Evaluate: x2 ex dx. Z Solution: We use integration by parts: u dv = u v v du. Z Z We have no inverse trig functions or logarithms, thus we choose u to be x2 and dv the leftover part:

u = x2 dv = ex dx

The left side, we di↵erentiate. The right side we integrate.

u = x2 dv = ex dx derivative 8 9 integral > & > <> du =2xdx v = ex =>

> > Now we plug into the formula: :> ;> x2 ex dx = x2 ex ex 2x dx Z Z = x2 ex 2 xex dx Z We need to use integration by parts again! Since we already did this one earlier, we just ﬁll in the answer: x2ex dx = x2 ex 2(xex ex )+C Z = x2 ex 2xex +2ex + C Math 267 (University of Calgary) Fall 2015 7 / 9 The last example generalizes. An integral such as xn ex dx would need to integrate by parts ntimes! Atablecanbeusedtoshortenthecalculationsinvolved(R shortcut method). Example

Evaluate: (x4 +3x2 + x +1)sinxdx. Z Solution: Typically, we need to use integration by parts four times! We use the shortcut method starting with u =(x4 +3x2 + x +1)anddv =sinxdx. We then di↵erentiate down the column corresponding to u until we hit zero. We integrate along down the column corresponding to dv. There is also a sign column starting with “+”andthenalternatingsignsthatwewilluselater. sign derivatives integrals + x4 + 3x2 + x + 1 sin x & 4x3 + 6x + 1 cos x & + 12x2 + 6 sin x 24x & cos x + 24 & sin x

0 & cos x Multiply along the diagonals shown in the table. In front of each product put the sign in the ﬁrst column: +(x4 + 3x2 + x + 1)( cos x) (4x3 + 6x + 1)( sin x)+(12x2 + 6)(cos x) (24x)(sin x)+(24)( cos x)+C (x4 +3x2 + x +1)sinxdx=(4x3 18x +1)sinx (x4 9x2 + x +19)cosx + C Z Math 267 (University of Calgary) Fall 2015 8 / 9 Example

Evaluate: ex sin xdx. Z Solution: Following LIATE we take u to be the trig function: u =sinx dv = ex dx derivative integral 8 & 9 <> du =cosxdx v = ex =>

Now we plug into the formula::> ;> ex sin xdx = ex sin x ex cos xdx Z Z It looks like we haven’t made any progress, but if we apply integration by parts again we get: u =cosx dv = ex dx derivative 8 9 integral > & > <> du = sin xdx v = ex => > > Now we plug into the formula::> ;> ex sin xdx = ex sin x ex cos x + ex sin xdx Z  Z Trick: We applied integration by parts twice and ended up with the SAME integral (up to a constant). We can then combine these common integrals together to get the answer. ex 2 ex sin xdx= ex sin x ex cos x ex sin xdx= (sin x cos x)+C ! 2 Z Z Math 267 (University of Calgary) Fall 2015 9 / 9