S=Integral xdx/sqrt(x-1). I have proceeded thus- Put sqrt(x-1)=u then x=u^2+1 and dx/sqrt(x-1)=2du. S=(u^2+1)2du/u =(2u+2/u)du=u^2+2 log u +C =(x-1)+ 2 log sqrt(x-1)=(x-1)+log(x-1)+C Required answer is 2/3*(x+2)sqrt(x-1)+C Have I proceeded correctly and if so can these 2 resul...
How to integrate dx/(4-5 sin x) using t-substitution method(i.e. taking tan x/2=t)?
Thanks. I checked and really got them as same. But can we analytically show them to be so? I tried but could not make it. Please guide.
s=çdx/(4+5cos x). By using t-substitution, i.e. t=tan(x/2) we get cos x=(1-t^2)/(1+t^2) and dx=2dt/(1+t^2). Substituting in s and simplifying, we get s= 2çdt/4(1+t^2)+5(1-t^2)=2çdt/(9-t^2). Using standard result çdx/(a^2-x^2)...
Book works out integral of sqtr(a^2-x^2)dx as a^2/2*arcsin(x/a)+x/2*sqrt(a^2-x^2) by taking x=a sin theta, and advises to work it out using x=a cos theta. I did and got the first term as -a^2/2*arccos(x/a)and the same second term. This means arcsin x/a=-arccos x/a which can be...
30x2+Ax1=(30+A)1.6 60+A=48+1.6A or 0.6A=12 or A=20. Total mix=50oz and cost=30*2+20=80$ Or 80/50=1.6$/oz.
3x-y=2 Multiplt it by 3 9x-3y=6 Add to it the other eq. 9x+3y=4 You get 18x=10 or x=5/9 Substitute in any eq. 3*5/9-y=2 or y=15/9-2=5/3 -2 or y=(5-6)/2=-1/6. Check: 3*5/9-(-1/6)=2 or 5/3+1/6=2 or (10+2)/6=2 ok. 9*5/9-3*1/3=4 or 5-1=4 ok.
Thank yoy very much Mr. Steve for making it so clear and easy.
I have done Int(1/ax)dx=1/a*Int(1/x)dx =[(log x)/a] + C. The working in the book is Int(1/ax)dx=1/a*Int(a/ax)dx =[(log ax)/a]+C. Which one is correct and where is the error in my working, if so?
v^2=u^2+2fs or f=(v^2-u^2)/2s =(7.4^2-79^2)/(2*710) =(54.76-6241)/1420=-4.36m/s^2 Negative acceleration (i.e. against the direction of motion)of 4.36m/s^2