Question

S(x^4-8x^3+4x)dx найти интегралы б) s(cos^2x sin x dx в) (e^3x+1)dx

Answer 1

$\int(x^4-8x^3+4x)dx=\frac{x^5}{5}-8\frac{x^4}{4}+4\frac{x^2}{2}+C=\frac{x^5}{5}-2x^4+2x^2+C\\\\\int cos^2x\cdot sinxdx=\int (cosx)^2\cdot (-d(cosx))=-\frac{cos^3x}{3}+C\\\\\int (e^{3x}+1)dx=\frac{1}{3}e^{3x}+x+C$

Answer 2

$a) \int{(x^4-8x^3+4x)}dx=\\ | \int{x^{\alpha}dx}= \frac{x^{\alpha+1}}{\alpha+1}+C|\\ = \int{x^4}dx-8\int{x^3}dx+4\int{x^1}dx=\\ = \frac{x^{4+1}}{4+1}-8 \frac{x^{3+1}}{3+1}+4 \frac{x^{1+1}}{1+1}+C=\\ = \frac{x^5}{5}- \frac{8x^4}{4}+ \frac{4x^2}{2}+C=\\ = \frac{x^5}{5}-2x^4+2x^2+c;\\$

$b) \int{\cos(2x)sin(x)}dx=|d(\cos(x))=-\sin(x)dx|=\\ =-\int{\cos(2x)d(\cos(x))}=\\ |\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha|\\ =-\int{(2\cos^2(x)-1)}d(\cos(x))=| t=\cos(x)|=\\ =-\int{(2t^2-1)}dt=|\int{x^{alpha}}dx= \frac{x^{\alpha+1}}{\alpha+1}+C|\\ =-2\int{t^2}dt+\int{t^0}dt=-2 \frac{t^{2+1}}{2+1}+ \frac{t^{0+1}}{0+1}=\\ =- \frac{2}{3}t^3+t+C=|t=\cos(x)|=\cos(x)- \frac{2}{3}\cos^3(x)+C=\\ \cos(x)(1- \frac{2}{3}\cos^2(x))+C=\\ =\cos(x)(1- \frac{2}{3}(1-\sin^2(x))+C=$
$=\cos(x)(1- \frac{2}{3}+ \frac{2}{3}\sin^2(x))+C=\\ =\cos(x)( \frac{1}{3}+ \frac{2}{3}\sin^2(x))+C=\\ = \frac{1}{3}\cos(x)(1+2\sin^2(x))+C;$

$c)\int(e^{3x}+1)dx=\int{e^{3x}}dx+\int{}dx=\\ |\int{e^x}dx=e^x+C; \int{x^\alpha}dx= \frac{x^{\alpha+1}}{\alpha+1}+C;d(x)= \frac{1}{3}dx|}\\ = \frac{1}{3}\int{e^{3x}}d(3x)+\int{x^0}dx=\\ = \frac{1}{3}e^{3x}+ \frac{x^{0+1}}{0+1}+C=\\ = \frac{1}{3}e^{3x}+x+C$