# Math Help - double integral

1. ## double integral

$\int_0^9 \int_0^{\sqrt{x}} y cos x^2 dy dx$
$cos x^2 \int_0^9 [\frac{y^2}{2}]_0^{\sqrt{x}} dx$
$\int_0^9 \frac{x}{2} cos x^2 dx$
$\frac{1}{2} \int_0^9 x cos x^2 dx$

let u = $x^2$
du = 2x dx
$\frac{1}{2}$ du = x dx

= $\frac{1}{4} \int_0^9 cos u du$
= $\frac{1}{4} - sin u ]_0^9$
= $- \frac{1}{4} sin 9$

$\int_0^9 \int_0^{\sqrt{x}} y cos x^2 dy dx$
$cos x^2 \int_0^9 [\frac{y^2}{2}]_0^{\sqrt{x}} dx$
$\int_0^9 \frac{x}{2} cos x^2 dx$
$\frac{1}{2} \int_0^9 x cos x^2 dx$

let u = $x^2$
du = 2x dx
$\frac{1}{2}$ du = x dx

= $\frac{1}{4} \int_0^9 cos u du$
= $\frac{1}{4} - sin u ]_0^9$
= $- \frac{1}{4} sin 9$

$\int_0^9 \int_0^{\sqrt{x}} y \times cos x^2 dy dx$

= $\int_0^9 \int_0^{\sqrt{x}} \frac{y^2}{2} \times cos x^2 dx$

= $\int_0^9 cos x^2 \times \frac{x}{2} dx$

let $u = x^2 \rightarrow du = 2x dx \rightarrow dx = \frac{du}{2x}$

the integral then becomes:

= $\int_0^9 cos u \times \frac{x}{2} \frac{du}{2x}$

= $\frac{1}{4}\int_0^9 cos u du$

= $\frac{1}{4} [sin(u)]_0^9$

= $\frac{1}{4} [sin(x^2)]_0^9$

= $\frac {sin(81)}{4}$

You did not substitute $u = x^2$ in the penultimate step of your solution!