$\int\sin\left(t^2\right)dt$

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Final answer to the problem

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nt^{\left(4n+3\right)}}{\left(4n+3\right)\left(2n+1\right)!}+C_0$
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Step-by-step Solution

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Apply the formula: $\sin\left(x^m\right)$$=\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\left(x^m\right)^{\left(2n+1\right)}$, where $x^m=t^2$, $x=t$ and $m=2$

$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\left(t^2\right)^{\left(2n+1\right)}dt$

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$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\left(t^2\right)^{\left(2n+1\right)}dt$

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Learn how to solve integrali trigonometrici problems step by step online. int(sin(t^2))dt. Apply the formula: \sin\left(x^m\right)=\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\left(x^m\right)^{\left(2n+1\right)}, where x^m=t^2, x=t and m=2. Simplify \left(t^2\right)^{\left(2n+1\right)} using the power of a power property: \left(a^m\right)^n=a^{m\cdot n}. In the expression, m equals 2 and n equals 2n+1. Apply the formula: x\left(a+b\right)=xa+xb, where a=2n, b=1, x=2 and a+b=2n+1. Apply the formula: \int\sum_{a}^{b} cxdx=\sum_{a}^{b} c\int xdx, where a=n=0, b=\infty , c=\frac{{\left(-1\right)}^n}{\left(2n+1\right)!} and x=t^{\left(4n+2\right)}.

Final answer to the problem

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nt^{\left(4n+3\right)}}{\left(4n+3\right)\left(2n+1\right)!}+C_0$

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Function Plot

Plotting: $\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nt^{\left(4n+3\right)}}{\left(4n+3\right)\left(2n+1\right)!}+C_0$

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1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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