# How do you integrate #sinh2xe^(cosh2x) dx#?

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To integrate ( \sinh(2x)e^{\cosh(2x)} ) with respect to ( x ), you can use integration by parts. Let ( u = \sinh(2x) ) and ( dv = e^{\cosh(2x)} dx ). Then, ( du = 2\cosh(2x) dx ) and ( v = e^{\cosh(2x)} ). Apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values:

[ \int \sinh(2x)e^{\cosh(2x)} dx = \sinh(2x)e^{\cosh(2x)} - \int e^{\cosh(2x)} \cdot 2\cosh(2x) dx ]

The integral on the right side is similar to the original integral. You can use integration by parts again:

[ \int e^{\cosh(2x)} \cdot 2\cosh(2x) dx ]

Let ( u = 2\cosh(2x) ) and ( dv = e^{\cosh(2x)} dx ). Then, ( du = 4\sinh(2x) dx ) and ( v = e^{\cosh(2x)} ). Apply integration by parts again:

[ \int e^{\cosh(2x)} \cdot 2\cosh(2x) dx = 2\cosh(2x)e^{\cosh(2x)} - \int 4\sinh(2x) e^{\cosh(2x)} dx ]

Now, you have another integral similar to the original one. You can continue this process until you reach an integral that you can evaluate.

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When evaluating a one-sided 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 one-sided 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 one-sided 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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