ACTIVITY: Evaluate each of the integrals listed below. Use integration by parts. Show your comprehensive solutions 1. 3xe2x dx 2. S e2x cos x dx 3. f In x dx 4. S x In 2x dx 5. S sec3xdx 6. fxsin2xdx 7. Stan 1 4x dxLESSON 15: TECHNIQUES - INTEGRATION BY PARTS X. Integration by Parts Objectives: 1. Convert integrand into something that can be solved using previously discussed theorems 2. Make the integration of complicated functions easier 3. Evaluate integrands containing the product of two mathematical functions, whether of the same type (i.e., both algebraic), or of different types (i.e., trigonometric and exponential, trigonometric and algebraic). 4. Reduce integrands of two mathematical functions into a single function. Recall: Differential of a Product: (uv) = u" + val - d(uv) = udv + vdu udv = d(uv) - vdu Integrating both sides, Sudv = ] d(uv) - fvdu Hence, T62] Sudv = uv - fvdu (Integration by Parts formula) ***This means that every given integrand in this topic, shall be treated as J udv Examples: Evaluate the following integrals: 1. [ xe*dx 2. x(2x + 5)13 dx 3. Prove T21 4. Stanh 1 xdx 5. J x2 cos 3x dxSolutions: 1. This is a product of an algebraic and an exponential function. Always your first step: Assign u and dv from the given problem Let u = x and dv = edx (why not u = ex and du=xdx?) From u and dv, obtain du and v du = dx, v= e* Apply T62, Judv = uv - fvdu -- Sxe*dx = xex - feldx (integrand reduced to a single function) = xex - ex + C 2. This is a product of two algebraic functions u = x, du = (2x + 5)"'dx - du = dx, v = (2x+5)12 24 = uv - fvdu = x(2x+5)12 (2x+5)12 -dx =*(2x+5) 12 (2x+5)13 24 24 24 -+ C 312 3. T21 states that [ sin 1 xdx = xsin 1x + v1-xz + c **This is a single mathematical function (inverse trigonometric) where integration by parts may also be applied.u = sin 'x, du = dx - du = V = X = uv - vdu = xsin ]x - = xsin 1 x - S(1 - x2)-1/2xdx (Power Rule) = x sin-1 x - (-)(1 -x2)1/2 + C = xsin 1 x+ V1 - x2 + C (T19 verified) 4. This is an integrand containing inverse hyperbolic function. This was not presented during our discussions on integration of transcendental functions, though we know its derivative. u = tanh x, du = dx - du = _ and v = x uv - fvdu = xtanh 1 x - fxdx (T5) xtanh-1 x - (-);S = xtanh-1 x + -In(1 -x?) + C "*You can derive the remaining formulas for integration of inverse hyperbolic functions using by parts. 5. u = x2, dv = cos 3x dx - du = 2xdx, v = = sin 3x = =x2 sin 3x -fx sin 3x dx (the integrand still has two functions, but x is now of lesser degree) For xsin3xdx: u = x, dv = sin 3xdx - du = dx, v= -=cos 3x5. u = x2, dv = cos 3xdx - du = 2xdx, v= = sin 3x =-x sin 3x - WIN [ x sin 3x dx (the integrand still has two functions, but x is now of lesser degree) For xsin3xdx: u = x, dv = sin 3xdx - du = dx, v=-=cos3x 3 = =x2 sin 3x - -}x cos 3x - (- S cos 3x dx) (integration by parts reapplied) = =x2 sin 3x + -x cos 3x- CIN [ cos 3x dx = WIH x sin 3x + =x cos 3x - 27 sin 3x + C