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

Joyce Smith 2022-01-07 Answered
Evaluate the indefinite integral.
\(\displaystyle\int{\frac{{{\cos{\sqrt{{{t}}}}}}}{{\sqrt{{{t}}}}}}\)

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Expert Answer

Maria Lopez
Answered 2022-01-08 Author has 4644 answers
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 \(\displaystyle{I}={2}{\sin{{\left(\sqrt{{{t}}}\right)}}}+{C}\).
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vicki331g8
Answered 2022-01-09 Author has 525 answers
\(\displaystyle\int{\frac{{{\cos{{\left(\sqrt{{{t}}}\right)}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}\)
put
\(\displaystyle\sqrt{{{t}}}={u}\)
\(\displaystyle{\frac{{{1}}}{{{2}\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={d}{u}\)
\(\displaystyle{\frac{{{1}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={2}{d}{u}\)
\(\displaystyle\int{\frac{{{\cos{{\left(\sqrt{{{t}}}\right)}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}=\int{\cos{{u}}}{2}{d}{u}={2}\int{\cos{{u}}}{d}{u}={2}{\sin{{u}}}+{C}\)
\(\displaystyle{u}=\sqrt{{{t}}}\)
\(\displaystyle\int{\frac{{{\cos{{\left(\sqrt{{{t}}}\right)}}}}}{{\sqrt{{{t}}}}}}{\left.{d}{t}\right.}={2}{\sin{\sqrt{{{t}}}}}+{C}\)
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karton
Answered 2022-01-11 Author has 8659 answers

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

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