Examples from Section 7.1: Integration by Parts Page 1

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Laureen Hodges
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1 Examples from Section 7.: Integration by Parts Page Questions Example Determine x cos x dx. Example e θ cos θ dθ Example You may wonder why we do not add a constant at the point where we integrate for v in the parts substitution. This is because any constant added there will cancel later. Redo x cos x dx and keep all constants of integration to show this. Example: / arccos x dx. Example Prove the following integral formula (# in table) using parts: [ ln w ] + c Solutions Example Determine x cos x dx. Let s do this integral using parts. The formula for parts is for u and dv based on our integral. Let s choose udv = uv vdu, so we want to choose the two substitutions u = x dv = cos x dx du = dx v = cos x dx = sin x NOTE: once you have chosen u, dv has to be everything else that s left so we can write: x cos x dx = u dv = x sin x sin x dx = x sin x ( cos x) + c (integrate) = x sin x + cos x + c (simplify) What we have done is replaced the original integral, which we could not do, with the integral do. This is the power of parts! sin x dx, which we could

2 Examples from Section 7.: Integration by Parts Page Check our answer by taking the derivative, and verifying that we get the integrand back: d [x sin x + cos x + c] = sin x + x cos x sin x = x cos x dx The problem most people have with parts is picking the u and dv correctly. Keep in mind that we want to be able to do our new integral, so generally we pick u to be a function that gets simpler when differentiated. What would happen if we had chosen differently in the above? Let s choose u = cos x dv = x dx du = sin x dx v = x dx = x x cos x dx = u dv = uv = x cos x + x sin x dx This is true, but it doesn t help us do the integral! Sometimes you have to use parts twice, and sometimes you never have to integrate explicitly to get the integral. Example e θ cos θ dθ We want to use parts, but neither e θ nor cos θ will simplify when we take a derivative. We have to do something, so let s try u = cos θ du = sin θ dθ dv = e θ dθ v = e θ dθ = e θ So the integral becomes: e θ cos θ dθ = u dv = e θ cos θ e θ ( sin θ) dθ = e θ cos θ + e θ sin θ dθ (simplify) () Our new integral is no simpler. But, it is no worse! That tells us to try parts again, on the the integral: e θ sin θ dθ

3 Examples from Section 7.: Integration by Parts Page 3 Let s choose u = sin θ du = cos θ dθ dv = e θ dθ v = e θ dθ = e θ NOTE: if we switch our choices of u and dv here, we will get the original integral back. So the integral becomes: e θ sin θ dθ = u dv = e θ sin θ e θ cos θ dθ Now, we can substitute everything back into Equation (): e θ cos θdθ = e θ cos θ + e θ sin θ dθ = e θ cos θ + e θ sin θ e θ cos θ dθ Solve for the integral algebraically: e θ cos θ dθ = [eθ cos θ + e θ sin θ] + c Notice how we simply added a constant of integration at the end, since we had a definite integral. Example You may wonder why we do not add a constant at the point where we integrate for v in the parts substitution. This is because any constant added there will cancel later. Let s redo x cos x dx to show this. Let s choose u = x dv = cos x dx du = dx v = cos x dx = sin x + c x cos x dx = u dv = x(sin x + c ) (sin x + c ) dx = x sin x + xc [( cos x) + c x] + c (integrate) = x sin x + cos x + c (simplify) Parts is also used to prove many of the integral formulas in the back of your textbook, and especially the ones that are written in terms of reduction formulas. The following example shows where #88 from the table of integrals comes from. Definite Integrals by Parts: b b u dv = uv b a. a a

4 Examples from Section 7.: Integration by Parts Page 4 Example: / arccos x dx. To begin, we use parts and choose: u = arccos x du = dx x dv = dx v = x / arccos x dx = / u dv = uv / / = x arccos x / x ( x ) dx = arccos arccos + / x x dx Substitution: t = x when x = t = dt = xdx when x = t = 3 4 = ( π ) ( π ) 3/4 dt + 3 t ( ) = π 6 = π 6 3/4 t / (/) t / dt 3/4 = π 6 ( 3/4 ) = π 6 + 3/4 Example Prove the following integral formula (# in table) using parts: [ ln w ] + c To begin, we use parts and choose: u = ln w du = w dw dv = w n dw v = w n dw = Notice that the above integral for v is only true if n. = u dv = uv

5 Examples from Section 7.: Integration by Parts Page 5 = ln w = ln w n + = ln w n + w dw w n dw ( w n+ w n+ n + = ln w + c = w n+ [ ln w ] + c ) + c (simplification) We have shown [ ln w ] + c, n. If n =, it is not too difficult to show ln w w dw = ln w + c.

Unit #11 : Integration by Parts, Average of a Function. Goals: Learning integration by parts. Computing the average value of a function.

Unit #11 : Integration by Parts, Average of a Function Goals: Learning integration by parts. Computing the average value of a function. Integration Method - By Parts - 1 Integration by Parts So far in