Answer :

Ideas required to solve the problems:


* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.


* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.


Let, I =


To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-


If I has the form


Then substitute numerator as -



Where A, B and C are constants


We have, I =


As I matches with the form described above, So we will take the steps as described.



{



Comparing both sides we have:


3B+ C = 3


B + 2A = 2


2B - A = 4


On solving for A ,B and C we have:


A = 0, B = 2 and C = -3


Thus I can be expressed as:


I =


I =


Let I1 = and I2 =


I = I1 + I2 ….equation 1


I1 =


So, I1 reduces to:


I1 = …..equation 2


As, I2 =


To solve the integrals of the form


To apply substitution method we take following procedure.


We substitute:



I2 =


I2 =


I2 =


I2 =


Let, t =


I2 =


As, the denominator is polynomial without any square root term. So one of the special integral will be used to solve I2.


I2 =


I2 =


I2 = { a2 + 2ab + b2 = (a+b)2}


As, I2 matches with the special integral form



I2 =


Putting value of t we have:


I2 = + C2 ……equation 3


From equation 1,2 and 3:


I = + C2


I = + C ….ans


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