Answer to Question #3467 in Calculus for noor

Question

What is the integration of

1- & & ( x[sup]2[/sup] + 1) cos 3x dx
2-& & & & сos ( x[sup]1/2[/sup] ) dx
3-& & & & e[sup]x[/sup] cos x d

Answer 1

<img src="/cgi-bin/mimetex.cgi?1.%20%5Cint%7B%281+x%5E2%29%5Ccos%7B3x%7D%20dx%7D%20=%20%5Cint%7B%5Ccos%7B3x%7Ddx%7D%20+%20%5Cint%7Bx%5E2%20%5Ccos%7B3x%7D%20dx%7D%20=%20%5C%5C%20%5Cbegin%7Bbmatrix%7D%20u%20=%20x%5E2%20&%20du%20=%202x%20dx%5C%5C%20dv%20=%20%5Ccos%7B3x%7Ddx%20&%20v%20=%20%5Cfrac%7B1%7D%7B3%7D%20%5Csin%7B3x%7D%20%5Cend%7Bbmatrix%7D%20=%20%5Cfrac%7B1%7D%7B3%7D%5Csin%7B3x%7D%20+%20%5Cfrac%7B1%7D%7B3%7Dx%5E2%5Csin%7B3x%7D%20-%20%5Cfrac%7B2%7D%7B3%7D%5Cint%7Bx%20%5Csin%7B3x%7D%7D%20=%20%5C%5C%20=[analog.]%20=%5C%5C%20=%20%5Cfrac%7B1%7D%7B3%7D%5Csin%7B3x%7D%281+x%5E2%29+%20%5Cfrac%7B2%7D%7B9%7Dx%5Ccos%7B3x%7D%20-%20%5Cfrac%7B2%7D%7B27%7D%5Csin%7B3x%7D%5C%5C%202.%20%5Cint%7B%5Ccos%7B%5Csqrt%7Bx%7D%7Ddx%7D%20=%20[%5Csqrt%7Bx%7D%20=%20t,%20dx%20=%202t%20dt]%20=%202%20%5Cint%7Bt%5Ccos%7Bt%7Ddt%7D%20=%20%5C%5C%20=%202t%20%5Csin%7Bt%7D%20-%202%5Cint%20%7B%5Csin%7Bt%7Ddt%7D=2t%20%5Csin%7Bt%7D%20+%202%5Ccos%7Bt%7D%20=%20%5C%5C%20=%202%20%5Csqrt%7Bx%7D%20%5Csin%7B%5Csqrt%7Bx%7D%7D%20+%202%20%5Ccos%7B%5Csqrt%7Bx%7D%7D%20%5C%5C%203.%20%5Cint%7Be%5Ex%20%5Ccos%7Bx%7D%20dx%7D%20=%20%5Cbegin%7Bbmatrix%7D%20e%5Ex%20=%20u%20&%20du%20=%20e%5Ex%20dx%5C%5C%20%5Ccos%7Bx%7D%20=%20dv%20&%20v%20=%20%5Csin%7Bx%7D%20%5Cend%7Bbmatrix%7D%20=%20e%5Ex%20%5Csin%7Bx%7D%20-%20%5Cint%7Be%5Ex%20%5Csin%7Bx%7D%20dx%7D%20=%20%5C%5C%20=%20e%5Ex%20%5C%20cos%7Bx%7D%20+%20e%5Ex%20%5Csin%7Bx%7D%20-%20%5Cint%7Be%5Ex%20%5Ccos%7Bx%7D%20dx%7D%5C%5C%202%5Cint%7Be%5Ex%20%5Ccos%7Bx%7D%20dx%7D%20=e%5Ex%20%28%5C%20cos%7Bx%7D%20+%20%5Csin%7Bx%7D%29%20%5C%5C%20%5Cint%7Be%5Ex%20%5Ccos%7Bx%7D%20dx%7D%20=%20%5Cfrac%7B1%7D%7B2%7De%5Ex%20%28%5C%20cos%7Bx%7D%20+%20%5Csin%7Bx%7D%29" title="1. \int{(1+x^2)\cos{3x} dx} = \int{\cos{3x}dx} + \int{x^2 \cos{3x} dx} = \\ \begin{bmatrix} u = x^2 & du = 2x dx\\ dv = \cos{3x}dx & v = \frac{1}{3} \sin{3x} \end{bmatrix} = \frac{1}{3}\sin{3x} + \frac{1}{3}x^2\sin{3x} - \frac{2}{3}\int{x \sin{3x}} = \\ =[analog.] =\\ = \frac{1}{3}\sin{3x}(1+x^2)+ \frac{2}{9}x\cos{3x} - \frac{2}{27}\sin{3x}\\ 2. \int{\cos{\sqrt{x}}dx} = [\sqrt{x} = t, dx = 2t dt] = 2 \int{t\cos{t}dt} = \\ = 2t \sin{t} - 2\int {\sin{t}dt}=2t \sin{t} + 2\cos{t} = \\ = 2 \sqrt{x} \sin{\sqrt{x}} + 2 \cos{\sqrt{x}} \\ 3. \int{e^x \cos{x} dx} = \begin{bmatrix} e^x = u & du = e^x dx\\ \cos{x} = dv & v = \sin{x} \end{bmatrix} = e^x \sin{x} - \int{e^x \sin{x} dx} = \\ = e^x \ cos{x} + e^x \sin{x} - \int{e^x \cos{x} dx}\\ 2\int{e^x \cos{x} dx} =e^x (\ cos{x} + \sin{x}) \\ \int{e^x \cos{x} dx} = \frac{1}{2}e^x (\ cos{x} + \sin{x})">