COURSE MODULES IN MATH 27 (ANALYTIC GEOMETRY AND CALCULUS II) MODULE 2 TECHNIQUES OF INTEGRATION

COURSE MODULES in
V V



MATH 27
V V



(Analytic Geometry and Calculus II)
V V V V




MODULE 2 V




TECHNIQUES OF INTEGRATION
V V




Lynie B. Dimasuay
V V



Author




Ben Paul B. Dela Cruz
V V V V


John Mark T. V V V


Lampos Arniel E.
V V V


Roxas V


Editors




Ariel L. Babierra V V


Wielson M. V V


Factolerin Gimelle B.
V V V


Gamilla Pierre
V V


Lance A. Tan V V V


Contributors

,UNITV2. V TECHNIQUESVOFVINTEGRATION

This Vunit Vwill Vfocus Von Vthe Vdifferent Vtechniques Vof Vintegration. V Some Vintegrand Vrequires Va
Vspecific Vapproach Vto Vbe Vable Vto Vevaluate Vthem. V So Vyou Vneed Vto V pay Vattention Vto Vthe Vgiven
Vintegrand Vso Vthat Vyou Vwill Vknow Vwhat Vtechniques Vto Vbe Vused. V But Vthis Vdoes Vnot Vmean Vthat Vthe

Vsimple Vsubstitution Vtechnique Vthat Vyou Vlearned Vin Vthe Vprevious Vunit Vwill Vno Vlonger Vwork Vhere. V In

Vfact, Vthat Vwill Valways Vbe Vyour Vfirst Voption Vbefore Vapplying Vthe Vother Vtechniques.



Our Vgoals Vfor Vthis Vunit Vare Vas Vfollows. V By Vthe Vend Vof Vthe Vunit Vyou Vshould Vbe Vable Vto

 perform Vintegration Vby Vparts;
 use Vtrigonometric Vsubstitution Vto Vevaluate Vsome Vintegral Vforms;
 decompose Vrational Vfunctions Vto Vpartial Vfractions;
 use Vproper Vsubstitute Vto Vevaluate Vsome Vintegral Vforms;
 determine Vwhether Van Vimproper Vintegral Vis Vconvergent Vor Vdivergent; Vand
 execute Vthe Vproper Vtechnique Vin Vevaluating Vintegrals.


REMINDER: V REVIEW Von Vintegral Vforms Vin VUNIT V1. VThese Vwill Vbe Vthe Vbasis Vof Vthe Vother Vsolvable
Vintegral Vforms Vin Vthis Vunit. VAlso, Vreview Vthe Vderivatives Vfor Vsolving Vdifferentials Vin Vcase Vsubstitution

Vwill Vbe Vused Vin Vsolving Vintegrals.



2.1 Integration Vby VParts

In Vthis Vsection, Vwe Vwill Vstudy Vone Vof Vthe Vmost Vimportant Vtechniques Vof Vintegration Vcalled
Vintegration Vby Vparts. VThis Vtechnique Vis Vapplicable Vto Vthe Vintegrand Vinvolving Vproducts Vof Valgebraic
Vand Vtranscendental Vfunctions Vor Vin Vsome Vcases, Vwhen Vthe Vintegrand Vis Va Vproduct Vof Vtranscendental

Vfunctions.

Here Vis Vthe Vforulation:

Let V u V and V v V be Vfunctions Vof V x V. VRecall Vthe Vproduct Vrule Vfor Vdifferentiation,

Dx VVuVVv V uVVDx VVv V vVVDx VVu VV, Vequivalently Vit Vcan Vbe Vexpresses Vas
V V



dVuVVv V uV  vVVdu V. V Taking Vthe Vintegrals Vof Vboth Vside, Vyields
V Vdv





 d VuVVv V V  V
 u 
 v VVdu

 dv
VV




 uVVv V  V
 u VVdv V  V
 v VVdu


 u VVdv V  V u V
 v VVdu

v
V V 

MUST VREMEMBER!!! VIntegration Vby Vparts V(IBP).


An Vintegral Vform V
 f VxVdx V can Vbe Vexpressed Vas V
 u VVdv V which Vis, Vin Vturn, Vequal Vto
MATH V27 V Lecture VGuide VUNIT V2




V





uVVv V V v VVdu V.
Once V u V and V dv V are Vdetermined, Vsolve V du V from V u V, Vand V dv Vfrom V v V. VThen, Vsolve Vthe
Vresulting Vform.
V(IMSP,UPLB)




 
f V x dx
V
V  V V
 u VVdv V 

uVVv V V v VVdu


2

, Some Vhelpful Vtips:
1. Try Vto Vlet V Vbe Va Vfunction Vwhose Vderivative Vis Va Vfunction Vsimpler Vthan V VThen V Vwill Vbe Vthe
Vremaining Vfactors Vof Vthe Vintegrand. VNote Vthat V V will Valways Vinclude Vthe V Vof Vthe Voriginal

Vintegrand.

2. Try Vto Vlet V V be Vthe Vmost Vcomplicated Vportion Vof Vthe Vintegrand Vthat Vfits Va Vbasic Vintegration
Vrule. VThen Vyour V Vwill Vbe Vthe Vremaining Vfactor(s) Vof Vthe Vintegrand.





Illustration V1. VUse VIBP Vto Vevaluate V x VcosVxdx V.
Solution:
Since V the V derivative V of V V will V result V to V a V simple V expression, V we V can V choose V V =V
V and V V=

��� V� V�� V so Vthat V�� V = V �� V and V� V = V ∫ V�� V = V ∫ V��� V� V�� V = V ��� V�.

Thus,


 x cos xdx xxsinsin xx  (cos
V V V V


V V
sin xdx
V V


x)  C
V
V


V
V


V
V


V V V


 VxVsin Vx V Vcos Vx V VC

Verify Vthat Vthe Vanswer Vis Vcorrect Vby Vshowing Vthat V ( V V V +V V + V ) V= V
V V V .





Illustration V2. VUse VIBP Vto Vevaluate V Arcsin Vxdx V.
Solution:
Note V that V our V only V choice V for V V here V is V V V and V V =V . V Thus, and V � V = V ∫ V��
√�−��
V V= V=


∫ V�� V = V �.
Thus,
dx

Show Vthat
 Arcsin Vxdx V VxVArcsin Vx V 1V
Vx
2


x. ∫�.
V

V
V
V
= V−√� V− V�� V+
√� V− V��




Therefore,
∫ V������ V� V �� V = V � V������ V� V+ V√� V− V�� V + V�.
MATH V27 V Lecture VGuide VUNIT V2




Verify Vthat Vthe Vanswer Vis Vcorrect Vby Vshowing Vthat V��(� V������ V� V+ V√� V− V�� V + V�) V =
V ������ V�.
V(IMSP,UPLB)




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