S shermaine New Member Apr 28, 2004 #1 Sorry about..... Juz got a little puzzled over sloving this integration....... Any one can help? sin^2 + cos^2tdt Pls advise...
Sorry about..... Juz got a little puzzled over sloving this integration....... Any one can help? sin^2 + cos^2tdt Pls advise...
Styx Active Member Apr 28, 2004 #2 shermaine said: Sorry about..... Juz got a little puzzled over sloving this integration....... Any one can help? sin^2 + cos^2tdt Pls advise... Click to expand... 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
shermaine said: Sorry about..... Juz got a little puzzled over sloving this integration....... Any one can help? sin^2 + cos^2tdt Pls advise... Click to expand... 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
F fat-tony Member Apr 28, 2004 #4 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
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
F fat-tony Member Apr 28, 2004 #5 shermaine said: Sorry..... The integration is sin^2t + cos^2tsintdt Click to expand... 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
shermaine said: Sorry..... The integration is sin^2t + cos^2tsintdt Click to expand... 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
Styx Active Member Apr 28, 2004 #7 a blooming good algebreac manipliation program (but cheating) a free prog is MuPad
Styx Active Member Apr 28, 2004 #8 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
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
S shermaine New Member Apr 30, 2004 #9 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
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
daviddoria New Member Apr 30, 2004 #10 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
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
Noggin Member May 18, 2004 #11 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)]])]....
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)]])]....