Explanation Given: ∫ 0 1 tan − 1 x ⋅ d x Key concepts applied: When we are finding integral of product of two functions we can make use of a special rule called integration by parts. This rule is derived from the product rule of derivative. ( u ⋅ v ) ' = u ⋅ v ' + v ⋅ u ' Product rule of differentiation , where u and v are two separate functions. ∫ ( u ⋅ v ) ' = ∫ u ⋅ v ' + ∫ v ⋅ u ' Integrating on both sides of the equation. We need not include a constant of integration on the left, since the integrals on the right will also have integration constants u v = ∫ u ⋅ v ' + ∫ v ⋅ u ' As integration reverses differentiation, we have that the integral and the derivative on the left side "cancel each other". u v − ∫ v ⋅ u ' = ∫ u ⋅ v ' + ∫ v ⋅ u ' − ∫ v ⋅ u ' u v − ∫ v ⋅ u ' = ∫ u ⋅ v ' + ∫ v ⋅ u ' − ∫ v ⋅ u ' Subtracting ∫ v ⋅ u ' on both sides and Simplifying the right side ∫ u ⋅ v ' = u v − ∫ v ⋅ u ' ⇒ ∫ u ⋅ d v = u v − ∫ v ⋅ d u We rewrite using the symmetric property of equality which states that, If a = b , then b = a . Choose u and d v in such a way that 33) u is easier to differentiate 34) d v is easier to integrate and doesn’t get complicated as we integrate it. 35) ∫ v ⋅ d u is easier to compute than ∫ u ⋅ d v 36) Sometimes it is necessary to integrate by parts more than once. A helpful rule of thumb is I LATE. Choose u based on which of these comes first: I: Inverse trigonometric functions (such as s i n − 1 ( x ) , c o s − 1 ( x ) , t a n − 1 ( x ) ) L: Logarithmic functions (such as l n ( x ) , l o g ( x ) ) A: Algebraic functions (such as x 2 , x 3 ) T: Trigonometric functions (such as s i n ( x ) , c o s ( x ) , t a n ( x ) ) E: Exponential functions (such as e x , 3 x ) This above formula helps us to work out a difficult integral into simpler one. Formula used: Common Differentiation Rules Common Integration Rules Common Integration Rules Integration by parts ( tan − 1 x ) ' = 1 1 + x 2 ⋅ d x ( x n ) ' = n ⋅ x n − 1 ⋅ d x ∫ c f ( u ) ⋅ d u = c ⋅ ∫ f ( u ) ⋅ d u ( c ) ' = 0 , where c any constant ∫ d u = u + c ∫ 1 u ⋅ d u = ln | u | + c ∫ a b u ⋅ d v = [ u ⋅ v ] a b − ∫ a b v ⋅ d u Calculation: Description ∫ 0 1 tan − 1 x ⋅ d x Step 1: Identify the u and d v of the given function using the ILATE rule. · Let d v be the most complicated portion of the integrand that can be easily integrated. · Let u be that part of the integrand whose derivative d u is a ‘simpler’ function than u itself...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 8.1, Problem 58E

To determine

To Compute:

The given definite integral using integration by parts

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Chapter 8.1, Problem 57E

Chapter 8.1, Problem 59E

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Chapter 8 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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The given definite integral using integration by parts

Solution Summary: The author explains integration by parts, which is derived from the product rule of derivative, where u and v are two separate functions.

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To Compute: The given definite integral using integration by parts

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Chapter 8 Solutions

To Compute:

The given definite integral using integration by parts