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Need Help In Solving Math

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shermaine

New Member
Sorry about.....
Juz got a little puzzled over sloving this integration.......
Any one can help?
sin^2 + cos^2tdt
Pls advise...
 
shermaine said:
Sorry about.....
Juz got a little puzzled over sloving this integration.......
Any one can help?
sin^2 + cos^2tdt
Pls advise...

if you mean
INT [ Sin^2(t) + Cos^2(t) ] dt

well

Cos(A+B) = CosACosB - SinASinB

thus

Cos(A-B) = CosACosB + SinASinB

Let A=B

Cos(A-A) = CosACosA + SinASinA
Cos(A-A) = Cos^2(A) + Sin^2(A)

Cos(0) = Cos^2(A) + Sin^2(A)
Thus

Cos(0) = 1

Thus

Int( 1) dt = t +C ; Where C is the integration constant

hence

INT [ Sin^2(t) + Cos^2(t) ] dt = t + C ; where C is the integration constant


I loved it when teachers gave silly questions like that and you think you did summing wrong

Have fun
 
If you don't want to remember all the derivation, just remember this:

sin(x)^2 + cos(x)^2 = 1

It's a trig identity!

From this, we can also derive a few others. Divide the entire thing by cos(x)^2:

tan(x)^2 + 1 = sec(x)^2.

Or, if instead we divide by sin(x)^2:

1 + cot(x)^2 = csc(x)^2
 
shermaine said:
Sorry.....
The integration is sin^2t + cos^2tsintdt

Oh. That gets uglier.

Maple says:

-(1/2)*cos(t)*sin(t) + t/2 - (1/3)*t*cos(t)^3 + (1/9)*cos(t)^2*sin(t) + 2/9*sin(t)

Maple has sometimes given me answers that were more complicated than necessary. I will say though, this is a crappy question :D
 
a blooming good algebreac manipliation program (but cheating)
a free prog is MuPad
 
Right:

for the function:

Sin^2(t) + Cos^2(t)Sin(t)
the integral w.r.t. t is:


-1/2*cos(t)sin(t) + 1/2t -1/3cos^3(t) + C
 
Hello,

I still dun understand how u guys get the integration for Cos^2(t)Sin(t)
? Could any one pls advise me?
Thanks Thanks
 
its called u substitution

int(cos^2(t)*sin(t))
let u=cos(t)
du=-sin(t)dt
-int(u^2 du)
-u^3/3
-cos(t)^3/3

hope this helps
 
Tabular method kicks the ever lovin crap out of the u substituion, ESPECIALLY when you have to go through multiple iterations of substitution.

When you have f(x) * g(x) that you have to integrate, if f(x) will differentiate to 0, then you do the following:

Code:
f(x)          g(x)
f'(x)         INT[g(x)]
f''(x)        INT[INT[g(x)]]
...
0             INT[INT[...[g(x)]]]

Then you alternate signs on the g(x)...

Code:
f(x)          +g(x)
f'(x)         -INT[g(x)]
f''(x)        +INT[INT[g(x)]]
...
0             -INT[INT[...[g(x)]]]

Then you take the first item of column one and multiply it by the SECOND item of column two, and add that to the second item of column 1 times the third item of column 2, plus column 1 item 3 times column 2 item 4....

(f(x) * -INT[g(x)]) + (f'(x) * INT[INT[g(x)]]) + (f''(x) * -INT[INT[INT[g(x)]])]....
 
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