Integration by parts.

The tuitorial entails solving integrals using integration by parts.This metod of integration requires the first step to identify u & dv from the integral and this is achieved by applying the rule rule of the thumb using the acronym;LIATE whose letters represent five functions to be encountered in the integrals.

The tuitorial has explained in detail the creteria of choosing u & dv and aslo various sample problems showig how the formula can be applied.

At the end of the tuitorial is an exercise where you are provided with the clue to get you started and the answer for the question.Kindly practice the questions for more understanding of the topic.Note that practice makes perfect,and it is essential in understanding mathematics which would translate to high performance.Kinly check the file attached.

Wish you a successful tuitorial session.

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    *********** by ******* *** integrate ******** ***** *********** ** ***** using *** ********* *** *** ** **** at *** guidelines *** selecting * *** dv-There *** always ********** *** these *** ********* *** *** ******* ********* ** * ***** **** ********* * ** ** *** ******** **** ***** ** **** **** ******* ********* ********** ******* **** functionA: Algebraic ********** ************* ********** *********** ************** ******* ********** ************ *** ******** In * is * *********** ******** *** x ** an algebraic function ********* ** let;u=In x *** dv = * ****** * ***** before * ** ******** * ** *** ** * ************ uv- * *** **** * ** ** * *** **** * **** * C = *** **** - ** * * ******************************************************************************** ******* ****** ************ *** equation In * is * *********** ******** and * ** an ********* function Therefore ** ********* dv * sin x (Since * comes ****** T in LIATE)du = ***** ** ** *** ** * ********** uv- = *** cos ** (-cos x) * ***** x) dx = **** * *** cos ** - ** * -cos * *** *** ** * * -cos **** cos ** + *** * * * ****************************************************** ******* ********** ************ ****** ** appears **** *********** ** ***** does not ***** but ***** * ******** **** ********* *** *** ********** ************ x2)-1dx ******* * * = ************ * *** * (- ******* * x **** ** ** * x2)1/2 ***** * * * * ** ** ** * ****** *** * * * * ** ** * ****** * C ******************************************************************************* general ********** *** ********* * *** **** *** u ** **** section ** *** ********* ***** ********** ** ** * ************* function **** * itself2 *** dv ** *** **** complicated ******* ** *** ********* that *** ** ********** ************ ******* ****** ** ** ********* dxSolutionThe ***** **** of thumb ******* ********** here ***** **** ** *** ** ********* *** algebraic functions ** ***** **** 2 of *** *********** ********** ***** *** **** that * (4 ********* ** *** **** ********* ******* ** the integrand **** *** be ****** ********** Therefore;dv * x (4 –x2)1/2 *** * * ** **** ********* factor ** *** ************** means **** ** * **** ***** ** ********* ** ** (4 ********* *** *** () (4 ********* = -(4 *********** *** *** ***** *** ********* (4 ********* ** * *** * (x2) **** –x2)3/2 ) - ** ********* **** dx * -x2(4 –x2)3/2 * (4 ********* () * * = -x2(4 ********* - (4 ********* + * **************************************************** repeated *********** of *********** ** partsOccasionally integration ** ***** *** ** ******** ** order ** *** *** ***** ******** *** ******* ****** ** ***** ** *** ************ the ******* *** * and ** ** the ********** applicationsSample ******* ********** ** *** * dxSolutionFrom the equation *** * ** * trigonometric function *** ** ** an ********* ******** ********* we ******** *** ** **** xdx ****** * ***** ****** * in ******** = ** ** *** v= * -cos ************ *** * ** = *** = ******* ** - * *** cos * ** Applying integration ** ***** for the ****** time ** ***** the ****** **** ** the ******** above) *** repeating *** same procedure of choosing u and dvu=x (Algebraic ********* *** ** * cos x (trigonometric ******************* *** dx and v== *** xx2 *** * dx = -x2 *** * ******** -x2 cos * +2(x *** *** -x2 *** * **** *** * * *** * * *** *** cos x *** *** * * **** * * * ***************************************************** is ********* to **** **** *** ******** ******** 2 **** *** original ******** should not be ************* problem ****** ** cos * ************** *** ******** ** is an *********** ******** *** *** * ** a ************* ********* ** ********* ** * ** (Since * ***** before E ** LIATE)du * -sin * ** and v=ex ** * exThereforeex *** * *** *** Cos * ** * ** ***** ** dxCos * ex + ** *** x ********** *********** ** ***** for *** ****** time ** ******** * ** * ex sin x ** the ****** part of *** integral above) and repeating *** **** procedure of ******** * *** **** *** x ************** ********* *** ** = ** (exponential ******************* *** *** * dx *** **** dx= **** *** * dx= *** * ex * ******* *** x *** Cos * ex * *** * - ** *** * dxThe ******** ******** appears on the RHS ** *** ********* therefore we collect like ***** ** ****** ** ** the *** ** follows;ex *** x ** + ex *** * *** ** *** x * sin x2ex cos * *** ** *** * * *** * * *** *** x dx * (ex *** * * *** x) * * ………………………………………………AnswerKinly ***** *** **** ************

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