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Mathematics: Post your doubts here!

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please answer it for me its killing me
i did this halfway only and attempting this wud break ma heart yet again... i think sumone else shud try it
il give u a hint... to find out perpendicular distance from one point to one plane we use the formula given in the question

where a b and c are where the point starts that is the coordinates of p... and x y and z wud be the normal coordinates
find two such equations one with m and one with n and equate them with each other
 
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in some question where equation comes down to 1.3^n=39 or zomething to solve it if we use logs will it given credit?
 
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i did this halfway only and attempting this wud break ma heart yet again... i think sumone else shud try it
il give u a hint... to find out perpendicular distance from one point to one plane we use the formula given in the question

where a b and c are where the point starts that is the coordinates of p... and x y and z wud be the normal coordinates
find two such equations one with m and one with n and equate them with each other
thanks it helped :)
 
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depends on u... i loved p3 as compared to s1.
ii gave accelerated... and u wont guess.... with s2
got a b :/
and that also after only two months study of s2 :(
 
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ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2

x= pi/4

so the limits be pi/4 and 0

by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx

now substituting

(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24
 
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