Methods of integration-1- integration by part

Methods of integration
I- Integration by parts
Sometimes we can recognize the differential to be integrated as a product of a function which is easily differentiated and a differential which is easily integrated. For example, if the problem is to find
??x^2 cos?x dx
We have two way to do this:
1- Tabular integration by parts
Which is used for
1) the products of polynomials and sine function, x^n sin??bx.?
2) the products of polynomials and cosine function, x^n cos?bx.
3) the products of polynomials and exponential function, ?x^n e?^ax.
In any of these three cases we choose the polynomial as u and the product of sine function and dx (cosine or exponential function and dx – respectively) as dv.
Example1: Evaluate
1.??x^3 cos?x dx 2.???x^2 e?^2x dx

??x^3 cos?x dx=x^3 sin?x+3x^2 cos?x-6x sin?x-6 cos?x+c



???x^2 e?^2x dx=( x^2 )/2 e^2x-( 2x )/4 e^2x+( 2 )/8 e^2x+c=( e^2x )/2 (x^2-x+( 1 )/2)+c

2- Integration by using the formula
???u dv=u v-???v du ??
Which is used for inverse trigonometric functions and logarithm functions.
Example2: Evaluate
1.??ln??x dx ? 2.??sin^(-1)?2x dx
3. ?_1^4??sec^(-1) ?x? dx
1. u=ln?x and dv=dx
du=dx/x and v=x
???u dv=u v-???v du ??
???ln??x dx = ? x ln?x ? –??x× dx/x
=x ln?x-??dx =x ln?x-x+c


2. u=sin^(-1)?2x and dv=dx
du=2dx/?(1-4x^2 ) and v=x
???u dv=u v-???v du ??
??sin^(-1)?2x dx=x sin^(-1)?2x-???2xdx/?(1-4x^2 ) ?
=x sin^(-1)?2x+( 1 )/2 ?(1-4x^2 )+c
3. u=sec^(-1) ?x and dv=dx
du=dx/(?x ?(x-1))×1/(2?x) =dx/(2x?(x-1)) and v=x
??_1^4??sec^(-1) ?x? dx=x? sec^(-1) ?x?|_1^4-?_1^4?xdx/(2x?(x-1))
=4sec^(-1) 2-sec^(-1) 1-?_1^4?dx/(2?(x-1))
=4×?/3-0-1/2 ? ?(x-1)×2?|_1^4
=4?/3-(?3-0)=4?/3- ?3


1.???x ln?x ? dx 2.???x^3 e^(-3x) ? dx
3.??tan^(-1)?x dx 4.?_0^1?sin^(-1)??x/?x dx
5.??(x^2+3x) sin?2xdx 6.?_0^(1??2)??2x sin^(-1)?(x^2 )dx ?