If u and v be two functions of x, then integral of product of these two functions is given by: 850_integration.JPG

Note: In applying the above rule care has to be taken in the selection of the first function(u) and the second function (v). Normally we use the following methods:

If in the product of the two functions, one of the functions is not directly integrable 
(e.g. lnx, sin-1x, cos-1x, tan-1x etc.) then we take it as the first function and the remaining function is taken as the second function. e.g. In the integration of 
òx tan-1x dx, tan-1x is taken as the first function and x as the second function.

If there is no other function, then unity is taken as the second function e.g. In the integration oftan-1x dx , tan-1x is taken as the first function and 1 as the second function.

If both of the function are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually we use the following preference order for the first function.

(Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)

In the above stated order, the function on the left is always chosen as the first function. This rule is called as ILATE e.g. In the integration of , x is taken as the first function and sinx is taken as the second function.

Example -11:            Evaluate 2475_integration.JPG.

Solution:       769_integration.JPG


Example -12: Evaluate ∫sec3 θdθ.

Solution:    I =  sec3 θdθ = secθ  sec2 θdθ –  tan θ(secθ tanθ) dθ

                         = secθ tanθ  ∫ (sec2 θ – 1) secθ dθ = secθtanθ – I + ln|secθ + tanθ|

                        => I = 1/2 [secθ tanθ] + 1/2 In |secθ tanθ| + c

Example -13: Evaluate ∫ x3 In xdx.

Solution: Here u = lnx => du = 1/x dx

                        and v = x3 => dv = 3x2 dx

                        I = 1/4 x4 In x – 1/4 x4 In x – 1/16 x4 + c

                        = x4 / 4 . In x – 1/4  x3 dx = 1/4 x4 In x – 1/6 x4 + c.

An important result: In the integral ∫g(x)exdx, if g(x) can be expressed as g(x) = f(x) + f'(x) then = ex [f(x) + f(x)] dx = exf(x) + c

Example -13: Evaluate  ex (1/x – 1/x2)dx.

Solution: Let I =  ex (1/x – 1/x2)dx

                        Here f(x) = 1/x, f'(x) = –1/x2

                        Hence I = ex / x + c 

Example -14: Evaluate  ex (1–x / 1+x2)2dx.


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