Partial Fractions of Decomposition - Assignment 3 | M 408C, Assignments of Mathematics

Download Partial Fractions of Decomposition - Assignment 3 | M 408C and more Assignments Mathematics in PDF only on Docsity! Solutions – M408C FER – Day 3 1. Determine which integration technique would be used to solve the following integration problems. Do not work the problem completely. A) 2 4 13 dx x x− +∫ (Complete the square, Inverse tangent function) B) 2 32 1 x dx x −∫ (U-substitution) C) 3 lnx x dx∫ (Integration by parts) D) 2 3 csc cot x dx x∫ (U-substitution) E) (Integration by parts, twice) 2 cos3xe x∫ dx F) 2 2 2 x dx x x−∫ (Complete the square, Trigonometric substitution) G) 3 2 2 6 5 s ds s−∫ (U-substitution) H) 2 4 8 x dx x x+ +∫ (Complete the square, Trigonometric substitution) I) 2 3 2 9 ( 2) x x dx x − −∫ (Partial fraction decomposition) J) 1 lne xdx x∫ (U-substitution) K) 1 2 xe dx x∫ (U-substitution) L) 4 0 cosx xdx π ∫ (Integration by parts) M) 2 2 4 7 ( 1)( 2 3) x x dx x x x − + + − +∫ (Partial fraction decomposition) N) 2 2xx e dx∫ (Integration by parts, twice) 2 3 2 2 3 3 2 2 csc5. cot Let cot csc 2 1 1 2 2cot x dx x u x du x du uu du C u C C u x − − = → = − − = − = − + − = + = + ∫ ∫ ∫ dx 2 2 2 2 2 2 2 1 1 2 2 6. Techniques: complete the square, forms of inverse trig functions 4 13 Note: 4 13 4 4 9 ( 2) 3 Let 3 and 2 ( 2) 3 1 1 2tan tan ( ) 3 3 dx x x x x x x x dx a u x du dx x du u xC C u a a a − − − + − + = − + + = − + = = − → = − + − = + = + + ∫ ∫ ∫ 1 1 2 2 7. 1 1Let x xu u e dx x u du dx x x e du e C e C = → = − − = − + = − ∫ ∫ + Integration by Parts Reminder: u dv uv v du= −∫ ∫ 3 4 3 4 4 3 4 4 3 4 4 8. ln 1let ln 4 1ln ln 4 4 ln 1 ln 1 4 4 4 4 4 ln 4 16 x x dx u x du dx x xdv x dx v x xx x dx x dx x x x x x x4x dx C x x x C = → = = → = = ⋅ − ⋅ ⎛ ⎞ = − = − ⋅ +⎜ ⎟ ⎝ ⎠ = − + ∫ ∫ ∫ ∫ 4 0 4 0 4 0 4 4 0 0 9. cos Let and cos sin cos sin sin 2 0 cos 4 2 2 2 2 4 21 8 2 8 x x dx u x du dx dv x dx v x 8 x x dx x x x dx x π π π π π π π π = → = = → = = − ⎛ ⎞ = ⋅ − +⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎛ ⎞ + − = + − =⎜ ⎟⎜ ⎟ ⎝ ⎠ ∫ ∫ ∫ 2 2 2 2 2 2 2 2 2 2 10. cos3 Let cos3 3sin 3 1 and 2 1 1cos3 cos3 ( 3sin 3 ) 2 2 1 3cos3 sin 3 2 2 Now, let sin 3 3cos3 1 and 2 3So the part 2 x x x x x x x x x x e x dx u x du x dx dv e dx v e e x dx x e e x dx e x e x dx u x du x dx dv e dx v e e = → = − = → = = ⋅ − ⋅ − = + = → = = → = ∫ ∫ ∫ ∫ 2 2 2 2 2 2 2 2 2 2 3 1 1sin 3 sin 3 3cos3 2 2 2 3 9sin 3 cos3 4 4 Putting it all together, we have: 1 3 9cos3 cos3 sin 3 cos3 2 4 4 9Now collect like terms by adding cos3 to 4 x x x x x x x x x x x dx x e e x dx e x e x dx e x dx e x e x e x dx e x dx ⎡ ⎤= ⋅ − ⋅⎢ ⎥⎣ ⎦ = − = + − ∫ ∫ ∫ ∫ ∫ ∫ 2 2 2 2 2 2 both sides. 13 1 3cos3 cos3 sin 3 4 2 4 4 1 3cos3 cos3 sin 3 13 2 4 x x x x x x e x dx e x e x e x dx e x e x C = + ⎡ ⎤= + +⎢ ⎥⎣ ⎦ ∫ ∫ 2 3 2 2 3 2 3 3 2 3 2 2 2 913. ( 2) We need to break this up using partial fractions. 2 9 4 4 2 ( 2) 2 ( 2) ( 2) ( 2) ( 4 ) (4 2 ) ( 2) 2 2 ( 4 ) 9 4 9 4 9 (4 2 x x dx x x x A B C Ax Ax A Bx B C x x x x x Ax A B x A B C x Ax x A A B x x A B B A A B − − − − + = + + = − − − − − + − + + − + = − = → = − + = − → − + = − → = − − + ∫ + − + 2 3 2 3 2 3 1 2 2 ) 0 4 2 2 1 10 2 9 2 1 10 ( 2) 2 ( 2) ( 2) 2 ln 2 ( 2) 10 ( 2) ( 2) 10( 2)2 ln 2 1 2 1 52ln 2 2 ( 2) C C A B A B C x x dx dx dx dx x x x x x x dx x dx x xx C x C x x − − − − = → = − + = = − = − − − − = + + − − − − = − − − − − − − = − − − + − − = − + + + − − ∫ ∫ ∫ ∫ ∫ ∫ 2 2 2 2 2 2 2 2 2 2 2 2 Trigonometric Substitution Form found in integral: substitution: by right triangle trigonometry: sin cos tan sec sec tan You a xa x a x a a xa x a x a x ax a a x a θ θ θ θ θ θ − − = = + + = = − − = = 2 2 2 2 2 2 can use u-substitution, if necessary, and have , ,a u a u x u− + − 2 2 2 2 2 2 2 2 2 2 2 14. 4 8 4 4 4 ( 2) 2 Form: Let tan and sec where 2 and 2 so 2 2 tan 2 tan 2 2sec Now substitute for and into the original integral: ( 2) 2 x dx x x x xdx dx x x x a u u a a u a u x a x x dx d x dx x dx x θ θ θ θ θ θ + + = = + + + + + + = + = = + = + = → = − → = + + ∫ ∫ ∫ ∫ 2 2 2 2 (2 tan 2)(2sec ) (2 tan 2)(sec ) 2sec 2 tan sec 2sec 2sec 2ln sec tan Now, substitute back in for the trigonometric functions: ( 2) 42tan ; sec ; 2 2 ( 2) 4 2= ( 2) 4 2ln 2 2 d d d C xx x xx C θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ − = = − = − = − + + + ++ = = + + + + + − + + ∫ ∫ ∫ θ