How do you find the definite integral for: $xcos(5x^2) dx$ for the intervals $[0,4sqrtpi]$ ?

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Answer 1 · 2016-04-01T12:18:51

Zero

$int_0^(4sqrtpi)xcos(5x^2)dx$

Use $x^2$ as your direct variable instead of $x$ :
$d(x^2)=2xdx$

Hence
$int_0^(4sqrtpi)xcos(5x^2)1/(2x)d(x^2)$
$1/2int_0^(4sqrtpi)cos(5x^2)d(x^2)$
$1/2sin(5x^2)$ from $0 to 4sqrtpi$

You can immediately see that the sine term will give zero value for both limits.

How do you find the definite integral for: xcos(5x^2) dxxcos(5x^2) dx for the intervals [0,4sqrtpi][0,4sqrtpi]?