# How do you find the antiderivative of #cos(x)sin(sin(x))dx#?

Substitute

Reverse the substitution to finish

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To find the antiderivative of ( \cos(x)\sin(\sin(x)) ) with respect to ( x ), you can use integration by parts.

Let: [ u = \cos(x) \Rightarrow du = -\sin(x)dx ] [ dv = \sin(\sin(x))dx \Rightarrow v = \int \sin(\sin(x))dx ]

Integrating ( v ) requires a substitution. Let: [ w = \sin(x) \Rightarrow dw = \cos(x)dx ]

This gives: [ v = \int \sin(w)dw = -\cos(w) + C = -\cos(\sin(x)) + C ]

Now, using the integration by parts formula: [ \int u dv = uv - \int v du ]

Substituting in our values: [ \int \cos(x)\sin(\sin(x))dx = -\cos(x)\cos(\sin(x)) - \int (-\cos(\sin(x)))(-\sin(x))dx ]

This simplifies to: [ \int \cos(x)\sin(\sin(x))dx = -\cos(x)\cos(\sin(x)) + \int \cos(\sin(x))\sin(x)dx ]

Now, integrate ( \int \cos(\sin(x))\sin(x)dx ): [ Let: \ z = \sin(x) \Rightarrow dz = \cos(x)dx ]

This gives: [ \int \cos(\sin(x))\sin(x)dx = \int z dz = \frac{1}{2}z^2 + C = \frac{1}{2}\sin^2(x) + C ]

Finally, substituting this back into our original equation: [ \int \cos(x)\sin(\sin(x))dx = -\cos(x)\cos(\sin(x)) + \frac{1}{2}\sin^2(x) + C ]

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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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