7.1 Integration by Parts - 7.1 Integration by Parts If u and v are functions of x and have continuous derivatives then u dv uv v du Note This is the

# 7.1 Integration by Parts - 7.1 Integration by Parts If u...

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7.1 Integration by Parts If u and v are functions of x and have continuous derivatives, then - = du v uv dv u Note: This is the formula for Integration by Parts
Guidelines for Integration by Parts Try letting dv be the most complicated portion of the integrand that fits a basic integration formula. Then u will be the remaining factor(s) of the integrand. Try letting u be the portion of the integrand whose derivative is a simpler function than u . Then dv will be the remaining factor(s) of the integrand.
Example 1 Evaluate the integral: - dx x ) ( sin 1 dx x du x u 2 1 1 1 ) ( sin let - = = - Solution: x v dx dv = = ( 29 C x x x C x x x dx x x x x dx x x x x dx x + - + = + - + = - - = - - = - - - - - - 2 1 2 / 1 2 1 2 / 1 2 1 2 1 1 1 ) ( sin 2 1 2 1 ) ( sin ) 1 ( ) ( sin 1 ) ( sin ) ( sin
Example 2- Repeated Application of Integration by Parts Evaluate the integral: - θ θ θ d e ) 2 cos( θ θ θ d du u ) 2 sin( 2 ) 2 cos( let - = = θ θ θ - - - = = e v d e dv - - - - - = θ θ θ θ θ θ θ θ d e e d e ) 2 sin( 2 ) 2 cos( ) 2 cos( θ θ θ d dU ) 2 cos( 4 ) 2sin(2 let U = = θ
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