Solve #\int(e^(2x))/(1+e^(4x))dx#, OR #\intx\arcsin(x)dx# using infinite series?
I have the solution using normal integration, but considering the rest of the worksheet problems revolve around using infinite series to solve integrals...
I have the solution using normal integration, but considering the rest of the worksheet problems revolve around using infinite series to solve integrals...
A)
B)
Using infinite Series.
Part (A)
We can form a power series for the integrand using the Maclaurin Series, so we have:
So, with:
Thus we get:
So we can write:
Part (B)
with:
Thus we get:
So we can write:
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To solve ( \int \frac{e^{2x}}{1+e^{4x}}dx ) using infinite series, we can expand ( \frac{1}{1+e^{4x}} ) as a geometric series and then integrate each term. Similarly, for ( \int x \arcsin(x)dx ), we'll express ( \arcsin(x) ) as its Maclaurin series and integrate the resulting series term by term.
For ( \int \frac{e^{2x}}{1+e^{4x}}dx ):

Expand ( \frac{1}{1+e^{4x}} ) as a geometric series: [ \frac{1}{1+e^{4x}} = \frac{1}{1+e^{2x}\cdot e^{2x}} = \frac{1}{1+e^{2x}}\cdot\frac{1}{1+e^{2x}} ] Expand ( \frac{1}{1+e^{2x}} ) using the geometric series formula: [ \frac{1}{1+e^{2x}} = \sum_{n=0}^{\infty} (1)^n e^{2nx} ]

Integrate each term of the series: [ \int e^{2nx} dx = \frac{1}{2n} e^{2nx} + C ]

Plug back the expression of the series and integrate term by term: [ \int \frac{e^{2x}}{1+e^{4x}}dx = \int \sum_{n=0}^{\infty} (1)^n e^{2nx} dx ] [ = \sum_{n=0}^{\infty} (1)^n \int e^{2nx} dx ]
For ( \int x \arcsin(x)dx ):

Express ( \arcsin(x) ) as its Maclaurin series: [ \arcsin(x) = \sum_{n=0}^{\infty} \frac{(1)^n}{(2n+1)2^n} x^{2n+1} ]

Integrate each term of the series: [ \int x^{2n+1} dx = \frac{1}{2n+2} x^{2n+2} + C ]

Plug back the expression of the series and integrate term by term: [ \int x \arcsin(x)dx = \int x \sum_{n=0}^{\infty} \frac{(1)^n}{(2n+1)2^n} x^{2n+1} dx ] [ = \sum_{n=0}^{\infty} \frac{(1)^n}{(2n+1)2^n} \int x^{2n+1} dx ]
Please note that both of these series representations may converge only for certain values of ( x ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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