INTEGRALS

A Note From now onwards, we shall write only one constant of integration in the final answer. (ii) We have x INTEGRALS (iii) We have e – e dx – – log + C – log + C Example Find the following integrals: (sin cos ) (ii) cosec (cosec cot ) (iii) Solution We have (sin cos ) = – cos We have (cosec (cosec + cot ) cosec cosec cot = – cot cosec x + (iii) We have dx – x dx – = tan x + Example Find the anti derivative F of f defined by f ( x ) = x – , where F ( ) = Solution One anti derivative of f ( x ) is x – x since ) x – x = x – Therefore, the anti derivative F is given by F( x ) = x – x + C, where C is constant. Given that F( ) = , which gives, = – × + C or C = Hence, the required anti derivative is the unique function F defined by F( x ) = x – x + . Remarks We see that if F is an anti derivative of f , then so is F + C, where C is any constant.

Thus, if we know one anti derivative F of a function f , we can write down an infinite number of anti derivatives of f by adding any constant to F expressed by F( x ) + C, C ∈ R . In applications, it is often necessary to satisfy an additional condition which then determines a specific value of C giving unique anti derivative of the given function. Sometimes, F is not expressible in terms of elementary functions viz., polynomial, logarithmic, exponential, trigonometric functions and their inverses etc. We are therefore blocked for finding .

For example, it is not possible to find – x by inspection since we can not find a function whose derivative is – x (iii) When the variable of integration is denoted by a variable other than x , the integral formulae are modified accordingly. For instance y y dy y EXERCISE . Find an anti derivative (or integral) of the following functions by the method of inspection. .

sin x . cos x . e x . ( ax + b ) .

sin x – e x Find the following integrals in Exercises to : . ( + ) . ( – . ax bx c dx .

. ( – x INTEGRALS . ( ) . ( 3cos .

( 3sin . (sec tan ) . cosec . – 3sin dx .

Choose the correct answer in Exercises and . . The anti derivative of equals (A) x (B) (C) x (D) . If d f x such that f ( ) = .

Then f ( x ) is (A) (B) (C) (D) . Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions. It was based on inspection, i.e., on the search of a function F whose derivative is f which led us to the integral of f . However, this method, which depends on inspection, is not very suitable for many functions.

Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms. Prominent among them are methods based on: . Integration by Substitution . Integration using Partial Fractions .

Integration by Parts . . Integration by substitution In this section, we consider the method of integration by substitution. The given integral can be transformed into another form by changing the independent variable x to t by substituting x = g ( t ).

Consider I = Put x = g ( t ) so that dx dt = g ′ ( t ). We write dx = g ′ ( t ) dt Thus I = ( ( )) f g t g t dt ′ This change of variable formula is one of the important tools available to us in the name of integration by substitution. It is often important to guess what will be the useful substitution. Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples.

Example Integrate the following functions w.r.t. x : sin mx x sin ( x + ) (iii) (iv) sin (tan – x Solution We know that derivative of mx is m . Thus, we make the substitution mx = t so that mdx = dt . mx dx t dt m = – m cos t + C = – m cos mx + C Derivative of x + is x .

Thus, we use the substitution x + = t so that x dx = dt . sin ( t dt = – cos t + C = – cos ( x + ) + C (iii) Derivative of x is . Thus, we use the substitution so that giving dx = t dt . Thus, tan t dt tan t dt Again, we make another substitution tan t = u so that sec t dt = du INTEGRALS tan t dt u du u + tan t + (since u = tan t ) tan C (since Hence, tan x + Alternatively , make the substitution tan (iv) Derivative of – x .

Thus, we use the substitution tan – x = t so that = dt. Therefore , sin (tan – x dx t dt = – cos t + C = – cos(tan – x ) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique. These will be used later without reference.