Joyce Smith

2022-01-07

Evaluate the indefinite integral.
$\int \frac{\mathrm{cos}\sqrt{t}}{\sqrt{t}}$

Maria Lopez

Expert

Step 1
We have the given integral as
$I=\int \frac{\mathrm{cos}\sqrt{t}}{\sqrt{t}}$
Let us consider that,
$\sqrt{t}=u$
$\frac{1}{2\sqrt{t}}dt=du$
$\frac{1}{\sqrt{t}}dt=2du$
Step 2
On substituting $\sqrt{t}=u$ and $\frac{1}{\sqrt{t}}dt=2du$ in our integral $I=\int \frac{\mathrm{cos}\sqrt{t}}{\sqrt{t}}dt$, we get the result as
$I=\int 2\mathrm{cos}\left(u\right)du$
$I=2\int \mathrm{cos}\left(u\right)du$
$I=2\mathrm{sin}u+C$
On substituting back $u=\sqrt{t}$, our integral becomes as
$I=2\mathrm{sin}\left(\sqrt{t}\right)+C$
Hence, value of $I=\int \frac{\mathrm{cos}\sqrt{t}}{\sqrt{t}}dt$ is $I=2\mathrm{sin}\left(\sqrt{t}\right)+C$.

vicki331g8

Expert

$\int \frac{\mathrm{cos}\left(\sqrt{t}\right)}{\sqrt{t}}dt$
put
$\sqrt{t}=u$
$\frac{1}{2\sqrt{t}}dt=du$
$\frac{1}{\sqrt{t}}dt=2du$
$\int \frac{\mathrm{cos}\left(\sqrt{t}\right)}{\sqrt{t}}dt=\int \mathrm{cos}u2du=2\int \mathrm{cos}udu=2\mathrm{sin}u+C$
$u=\sqrt{t}$
$\int \frac{\mathrm{cos}\left(\sqrt{t}\right)}{\sqrt{t}}dt=2\mathrm{sin}\sqrt{t}+C$

karton

Expert

$\int \frac{\mathrm{cos}\left(\sqrt{t}\right)}{\sqrt{t}}dt\phantom{\rule{0ex}{0ex}}\int 2\mathrm{cos}\left(u\right)du\phantom{\rule{0ex}{0ex}}2×\int \mathrm{cos}\left(u\right)du\phantom{\rule{0ex}{0ex}}2\mathrm{sin}\left(u\right)\phantom{\rule{0ex}{0ex}}2\mathrm{sin}\left(\sqrt{t}\right)\phantom{\rule{0ex}{0ex}}Solution:\phantom{\rule{0ex}{0ex}}2\mathrm{sin}\left(\sqrt{t}\right)+C$