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i changed it to 2 integral whatever limit 0 2 and cbf subbing in your limits as well :x

i wna see if my answer is right \int_{-2}^{2} x^2sinxdx i did integration by parts twice, but my answer seems wrong still cos i've never seen sin2, cos2 o-o i got 8sin2 -4cos2 - 4

Sketch the region cut off in the first quadrant by y = 2x-3 / (x^2 - 3x -4) and find its area. weird question o-o you get ln {-} which you cant do, both limits end up with with - ln but the answer you get is just positive ln of the two O_O

how do you know the height and width of the rectangles?

Use upper and lower rectangles to prove that \frac {1}{2} < \int_{2^n}^{2^{n+1}}\frac{1}{x} dx <1 , for\ n \geq 0 how do you do upper lower rectangles o-o

how about if it says "prove". Then you must go from LHS to RHS?

i remember my maths teacher said you always have to go from LHS = RHS. If you do RHS = LHS you get zero, when marked by strict HSC markers.

show that 1/ (x^2 -9) = 1/6 (1/(x-3) - 1/(x+3)) i . e \frac {1}{x^2-9} = \frac {1}{6}(\frac {1}{x-3} - \frac {1}{x+3}) you cant go backwards right

given that cos3x = 4cos^3x - 3cosx and sin3x = 3sinx - 4sin^3x, find S dx/ [(4x^3-3x)root(1-x^2)] i have tried and failed this ><

i dont have the answers and only semi learnt this so gta guess is the answer \frac {-x}{e^x} cos e^x

\int e^x sine^xdx\\ yes\ another\ question\ :)

is the answer 2ln root 3 -1?? btw how do you know what to substitute if it doesnt tell you for trig and please help with the show sin one :P

says show so you cant just draw it. i can't remember doubl angles :D i need help with : Use the substitution x = 2sin@ to evaluate integral between 1 and 0 x^2/ [root(9-x^2)]^3 dx \int_{0}^{1}\frac {x^2}{\sqrt {4-x^2}}dx

show that sin@ = 2t / 1+t^2

lol @ me i shouldnt do maths at night xD

\int \frac {cos\theta} {sin^5\theta} d\theta

integrate cos@/sin^5@ d@

woot i got it right xD

integrate sin5xcos3x dx need someone to check answer xD mines prob bs i got -cos8x/16 - cos2x/4 + c

i got it a bit after i posted xD can you do root (x^2-9) ^-1/2 and then integrate normally? is it still correct?