Question

Интегралы!!! помогите

Answer 1

$a)intlimits^2_1 {(4x+3-frac{4}{x^2})} , dx=frac{4x^2}{2}+3x-frac{4*x^{-1}}{-1}=(2x^2+3x+frac{4}{x})|limits^2_1=\=(8+6-1)-(2+3+4)=4$

 

$b)intlimits^4_1{(frac{sqrt{x}}{x}+8(2x-5)^3)dx}=intlimits^4_1{(frac{sqrt{x}}{sqrt{x}sqrt{x}}+8(2x-5)^3)dx}=\=intlimits^4_1{(frac{1}{sqrt{x}}+8(2x-5)^3)dx}=frac{x^frac{1}{2}}{frac{1}{2}}+frac{8(2x-5)^4}{4*2}=\=(2sqrt{x}+(2x-5)^4)|limits^4_1=(2*2+81)-(2+81)=\=4+81-2-81=2$

 

$c)intlimits^{-frac{pi}{4}}_{-frac{pi}{2}}{frac{dx}{cos^2x-1}} = intlimits^{-frac{pi}{4}}_{-frac{pi}{2}}{frac{dx}{1-sin^2x-1}} = intlimits^{-frac{pi}{4}}_{-frac{pi}{2}}{-frac{dx}{sin^2x}} = -(-ctgx)=\=ctgx|limits^{-frac{pi}{4}}_{-frac{pi}{2}}=ctg(-frac{pi}{4})-ctg(-frac{pi}{2})=-1-0=-1$

 

$d)intlimits^{2pi}_0{(cosfrac{x}{8}-sinfrac{x}{8})^2}dx=d)intlimits^{2pi}_0{(cos^2frac{x}{8}-2cosfrac{x}{8}sinfrac{x}{8}+sin^2frac{x}{8})}dx=\ =intlimits^{2pi}_0{(1-sinfrac{x}{4})}dx=(x+4cosfrac{x}{4})|limits^{2pi}_0=(2pi+4cosfrac{pi}{2})-\-(0+4cos0)=2pi-4$