1.\(\int x^2\sqrt{2-x} dx\)
2.\(\int xsec^22x dx\)
3) \(\int x/\sqrt{x+5} dx\)
+118723 1. integral of x^2(2-x)^0.5 dx
\(\int\;x^2\sqrt{2-x}\;dx\\~\\ \)
2-x has to be positive so I am going to substitute it for y^2
\(2-x=y^2\\ 2-y^2=x\\ x=2-y^2\\ \frac{dx}{dy}=-2y\\ dx=2y\;dy \)
\(\int\;x^2\;\sqrt{2-x}\;dx\\ =\int \;(2-y^2)^2*\sqrt{y^2}\;*2y\;dy\\ =\int \;(4-4y^2+y^4)*y\;*2y\;dy\\ =\int \;2y^2(4-4y^2+y^4)\;dy\\ =\int \;(8y^2-8y^4+2y^6)\;dy\\ =\frac{8y^3}{3}-\frac{8y^5}{5}+\frac{2y^7}{7}+c\\~\\ =\frac{8(2-x)^{1.5}}{3}-\frac{8(2-x)^{2.5}}{5}+\frac{2(2-x)^{3.5}}{7}+c\\~\\\)
I haven't had much practice with these but that looks alright :)
I think so anyway.
+118723 I've made a slight error here.
I can't edit so I will just explain.
dx=-2y dy
I forgot the negative sign so the answer should all be in brackets with a minus sign out the front.
I might as well simplify the answer while I am at it :)
\(\int\;x^2\;\sqrt{2-x}\;dx\\ =-\left[ \frac{8(2-x)^{1.5}}{3}-\frac{8(2-x)^{2.5}}{5}+\frac{2(2-x)^{3.5}}{7}\right]+c\\ =-\frac{2}{3*5*7}\left[ \frac{5*7*4(2-x)^{1.5}}{1}-\frac{3*7*4(2-x)^{2.5}}{1}+\frac{3*5*(2-x)^{3.5}}{1}\right]+c\\ =-\frac{2}{105}\left[ \frac{140(2-x)^{1.5}}{1}-\frac{84(2-x)^{2.5}}{1}+\frac{15(2-x)^{3.5}}{1}\right]+c\\ =-\frac{2}{140}\left[ 140(2-x)^{1.5}-84(2-x)^{2.5}+15(2-x)^{3.5}\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-84(2-x)+15(2-x)^{2}\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-168+84x+15(4-4x+x^2)\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-168+84x+60-60x+15x^2\right]+c\\ =\frac{-2(2-x)^{1.5}}{140}\left[ 15x^2+24x+32 \right]+c\\\)
+118723 2) \(\int x(sec^2(2x))\;dx\)
I did this with integration by parts.
\(\boxed{\int v \frac{du}{dx}dx=uv-\int u\frac{dv}{dx}dx}\\~\\ v=x \qquad \frac{du}{dx}=sec^2(2x)\\ \frac{dv}{dx}=1 \qquad u=\frac{tan(2x)}{2} \\~\\ \int\;x(sec^2(2x))\;dx\\ =\frac{xtan(2x)}{2}-\int\;1*\frac{tan(2x)}{2}\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\int tan(2x)\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\int \frac{sin(2x)}{cos(2x)}\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\frac{ln(cos(2x)}{-2}+c\\ =\frac{xtan(2x)}{2}+\frac{ln(cos(2x))}{4}+c\\\)
+118723 3)
\(\int\frac{x}{\sqrt{x+5}}\;dx\\ let\;\;u=x+5\\ so\;\;x=u-5\\ and\;\;dx=du\\~\\ =\int\;\frac{u-5}{\sqrt u}\;du\\ =\int\;u^{0.5}-5u^{-0.5}\;du\\ =\frac{u^{1.5}}{1.5}-\frac{5u^{0.5}}{0.5}+c\\ =\frac{2u^{1.5}}{3}-\frac{10u^{0.5}}{1}+c\\ =\frac{2u^{1.5}}{3}-\frac{30u^{0.5}}{3}+c\\ =\frac{2u^{0.5}}{3}\left[{u}-15\right]+c\\ =\frac{2\sqrt{x+5}}{3}\left[x+5-15\right]+c\\ =\frac{2\sqrt{x+5}}{3}\left[x-10\right]+c\\ =\frac{2\sqrt{x+5}\;(x-10)}{3}+c\\ \)
1=
Take the integral:
integral sqrt(2-x) x^2 dx
For the integrand sqrt(2-x) x^2, substitute u = sqrt(2-x) and du = -1/(2 sqrt(2-x)) dx:
= -2 integral u^2 (2-u^2)^2 du
Expanding the integrand u^2 (2-u^2)^2 gives u^6-4 u^4+4 u^2:
= -2 integral (u^6-4 u^4+4 u^2) du
Integrate the sum term by term and factor out constants:
= -2 integral u^6 du+8 integral u^4 du-8 integral u^2 du
The integral of u^6 is u^7/7:
= -(2 u^7)/7+8 integral u^4 du-8 integral u^2 du
The integral of u^4 is u^5/5:
= (8 u^5)/5-(2 u^7)/7-8 integral u^2 du
The integral of u^2 is u^3/3:
= -(2 u^7)/7+(8 u^5)/5-(8 u^3)/3+constant
Substitute back for u = sqrt(2-x):
= -2/7 (2-x)^(7/2)+8/5 (2-x)^(5/2)-8/3 (2-x)^(3/2)+constant
Which is equal to:
Answer: | = -2/105 (2-x)^(3/2) (15 x^2+24 x+32)+constant
2-
integral x sec^2(x) 2 x dx = 2 (x (-i x+2 log(1+e^(2 i x))+x tan(x))-i Li_2(-e^(2 i x)))+constant
3-
Take the integral:
integral x/sqrt(x+5) dx
For the integrand x/sqrt(x+5), substitute u = x+5 and du = dx:
= integral (u-5)/sqrt(u) du
Expanding the integrand (u-5)/sqrt(u) gives sqrt(u)-5/sqrt(u):
= integral (sqrt(u)-5/sqrt(u)) du
Integrate the sum term by term and factor out constants:
= integral sqrt(u) du-5 integral 1/sqrt(u) du
The integral of sqrt(u) is (2 u^(3/2))/3:
= (2 u^(3/2))/3-5 integral 1/sqrt(u) du
The integral of 1/sqrt(u) is 2 sqrt(u):
= (2 u^(3/2))/3-10 sqrt(u)+constant
Substitute back for u = x+5:
= 2/3 (x+5)^(3/2)-10 sqrt(x+5)+constant
Which is equal to:
Answer: | = 2/3 (x-10) sqrt(x+5)+constant
