What is the integral of $x^{3} cos (x^{2}) d x$ $x^3 cos(x^2) dx$ ?

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What is the integral of $x^{3} cos (x^{2}) d x$ $x^3 cos(x^2) dx$ ?

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Answer 1 · 2016-03-21T17:34:08

$\frac{1}{2} x^{2} sin (x^{2}) + \frac{1}{2} cos (x^{2}) + C$ $1/2x^2sin(x^2)+1/2cos(x^2) +C$

Explanation:

We can't just integrate straight away, so we try substitution.

While trying substitution, we observe that we could integrate $cos (x^{2}) x d x$ $cos(x^2) x dx$ by substitution.

So, let's split the integrand and use integration by parts.

$\int x^{3} cos (x^{2}) d x = \int x^{2} cos (x^{2}) x d x$ $int x^3 cos(x^2) dx = int x^2 cos(x^2) x dx$

(We notice that we could rewrite this as $\int u cos (u) \frac{1}{2} d u$ $int u cos(u) 1/2du$ , but we don't see how to integrate that, so we'll continue with parts for now.)

$\int x^{2} cos (x^{2}) x d x$ $int x^2 cos(x^2) x dx$

Let $u = x^{2}$ $u = x^2$ and $d v = cos (x^{2}) x d x$ $dv = cos(x^2) x dx$ .

Clearly $d u = 2 x d x$ $du = 2x dx$ , and

we can integrate $d v$ $dv$ by substitution to get.

$\frac{1}{2} sin (x^{2})$ $1/2sin(x^2)$ .

$u v - \int v d u = \frac{1}{2} x^{2} sin (x^{2}) - \int x sin (x^{2}) d x$ $uv-int vdu = 1/2x^2sin(x^2)-intxsin(x^2)dx$

Integrate by substitution agan to finish.

$\int x^{3} cos (x^{2}) d x = \frac{1}{2} x^{2} sin (x^{2}) + \frac{1}{2} cos (x^{2}) + C$ $int x^3 cos(x^2) dx = 1/2x^2sin(x^2)+1/2cos(x^2) +C$

Check the answer by differentiating.

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What is the integral of $x^{3} cos (x^{2}) d x$ $x^3 cos(x^2) dx$ ?

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