What is the antiderivative of $cos (x) sin (sin (x)) d x$ $cos(x)sin(sin(x))dx$ ?

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What is the antiderivative of $cos (x) sin (sin (x)) d x$ $cos(x)sin(sin(x))dx$ ?

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Answer 1 · 2016-04-19T03:03:18

$- cos (sin (x)) + C$ $-cos(sin(x))+C$

Explanation:

We will want to make use of the following integral:

$\int sin (u) d u = - cos (u) + C$ $intsin(u)du=-cos(u)+C$

Thus, we want to set the interior of the $sin$ $sin$ function, which is in fact another $sin$ $sin$ function, equal to $u$ $u$ .

So, we want to set $u = sin (x)$ $u=sin(x)$ . This implies that $\frac{d u}{d x} = cos (x)$ $(du)/dx=cos(x)$ and $d u = cos (x) d x$ $du=cos(x)dx$ .

Thus, we have

$\int cos (x) sin (sin (x)) d x = \int sin (sin (x)) \cdot cos (x) d x$ $intcos(x)sin(sin(x))dx=intsin(sin(x))*cos(x)dx$

$= \int sin (u) d u = - cos (u) + C = - cos (sin (x)) + C$ $=intsin(u)du=-cos(u)+C=-cos(sin(x))+C$

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What is the antiderivative of $cos (x) sin (sin (x)) d x$ $cos(x)sin(sin(x))dx$ ?

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Explanation:

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What is the antiderivative of $cos (x) sin (sin (x)) d x$ $cos(x)sin(sin(x))dx$ ?