help me integrate these
sin(x^2)dx
plz the answer is in infinte series
http://www.wolframalpha.com/input/?i=solve+dy+%3D+sin(x%5E2)+dx
If y = ∑an x^n then
y' = ∑n an x^(n-1)
y' = 2x cos(x^2)
You know the power series for cos(u), so expand both and equate coefficients to get a recurrence relation.
google can show you lots of examples of solution by power series
To integrate sin(x^2)dx using an infinite series, we can use the Taylor series expansion of sin(x).
The Taylor series expansion of sin(x) is given by:
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
Let's substitute x^2 into the Taylor series expansion of sin(x):
sin(x^2) = (x^2) - ((x^2)^3)/3! + ((x^2)^5)/5! - ((x^2)^7)/7! + ...
simplifying the expression:
sin(x^2) = x^2 - (x^6)/3! + (x^10)/5! - (x^14)/7! + ...
Now, let's integrate sin(x^2) by integrating each term separately:
∫ sin(x^2) dx = ∫ (x^2) dx - ∫ (x^6)/3! dx + ∫ (x^10)/5! dx - ∫ (x^14)/7! dx + ...
The integral of each term gives:
∫ (x^2) dx = (1/3) x^3 + C1 (integration constant)
∫ (x^6)/3! dx = (1/7) x^7 + C2
∫ (x^10)/5! dx = (1/11) x^11 + C3
∫ (x^14)/7! dx = (1/15) x^15 + C4
Combining all the terms, we get:
∫ sin(x^2) dx = (1/3) x^3 - (1/7) x^7/3! + (1/11) x^11/5! - (1/15) x^15/7! + C
where C represents the integration constant.
So, the integral of sin(x^2)dx as an infinite series is:
(1/3) x^3 - (1/7) x^7/3! + (1/11) x^11/5! - (1/15) x^15/7! + ...
Please note that this is an infinite series approximation of the integral and may not represent the exact value of the integral.