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

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

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Step 1
We have the given integral as
$$\displaystyle{I}=\int{\frac{{{\cos{\sqrt{{{t}}}}}}}{{\sqrt{{{t}}}}}}$$
Let us consider that,
$$\displaystyle\sqrt{{{t}}}={u}$$
$$\displaystyle{\frac{{{1}}}{{{2}\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={d}{u}$$
$$\displaystyle{\frac{{{1}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={2}{d}{u}$$
Step 2
On substituting $$\displaystyle\sqrt{{{t}}}={u}$$ and $$\displaystyle{\frac{{{1}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={2}{d}{u}$$ in our integral $$\displaystyle{I}=\int{\frac{{{\cos{\sqrt{{{t}}}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}$$, we get the result as
$$\displaystyle{I}=\int{2}{\cos{{\left({u}\right)}}}{d}{u}$$
$$\displaystyle{I}={2}\int{\cos{{\left({u}\right)}}}{d}{u}$$
$$\displaystyle{I}={2}{\sin{{u}}}+{C}$$
On substituting back $$\displaystyle{u}=\sqrt{{{t}}}$$, our integral becomes as
$$\displaystyle{I}={2}{\sin{{\left(\sqrt{{{t}}}\right)}}}+{C}$$
Hence, value of $$\displaystyle{I}=\int{\frac{{{\cos{\sqrt{{{t}}}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}$$ is $$\displaystyle{I}={2}{\sin{{\left(\sqrt{{{t}}}\right)}}}+{C}$$.
###### Not exactly what youâ€™re looking for?
vicki331g8
$$\displaystyle\int{\frac{{{\cos{{\left(\sqrt{{{t}}}\right)}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}$$
put
$$\displaystyle\sqrt{{{t}}}={u}$$
$$\displaystyle{\frac{{{1}}}{{{2}\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={d}{u}$$
$$\displaystyle{\frac{{{1}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={2}{d}{u}$$
$$\displaystyle\int{\frac{{{\cos{{\left(\sqrt{{{t}}}\right)}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}=\int{\cos{{u}}}{2}{d}{u}={2}\int{\cos{{u}}}{d}{u}={2}{\sin{{u}}}+{C}$$
$$\displaystyle{u}=\sqrt{{{t}}}$$
$$\displaystyle\int{\frac{{{\cos{{\left(\sqrt{{{t}}}\right)}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={2}{\sin{\sqrt{{{t}}}}}+{C}$$
karton

$$\int \frac{\cos (\sqrt{t})}{\sqrt{t}}dt \\\int 2\cos (u)du \\2\times \int \cos(u)du \\2\sin (u) \\2\sin (\sqrt{t}) \\Solution: \\2\sin (\sqrt{t})+C$$