Answered: "Evaluate the indefinite integral. ∫cos⁡tt"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Joyce Smith

Answered question

2022-01-07

Evaluate the indefinite integral.

\int \frac{\cos \sqrt{t}}{\sqrt{t}}

Answer & Explanation

Maria Lopez

Beginner2022-01-08Added 32 answers

Step 1
We have the given integral as

I = \int \frac{\cos \sqrt{t}}{\sqrt{t}}

Let us consider that,

\sqrt{t} = u

\frac{1}{2 \sqrt{t}} d t = d u

\frac{1}{\sqrt{t}} d t = 2 d u

Step 2
On substituting

\sqrt{t} = u

and

\frac{1}{\sqrt{t}} d t = 2 d u

in our integral

I = \int \frac{\cos \sqrt{t}}{\sqrt{t}} d t

, we get the result as

I = \int 2 \cos (u) d u

I = 2 \int \cos (u) d u

I = 2 \sin u + C

On substituting back

u = \sqrt{t}

, our integral becomes as

I = 2 \sin (\sqrt{t}) + C

Hence, value of

I = \int \frac{\cos \sqrt{t}}{\sqrt{t}} d t

I = 2 \sin (\sqrt{t}) + C

vicki331g8

Beginner2022-01-09Added 37 answers

\int \frac{\cos (\sqrt{t})}{\sqrt{t}} d t

put

\sqrt{t} = u

\frac{1}{2 \sqrt{t}} d t = d u

\frac{1}{\sqrt{t}} d t = 2 d u

\int \frac{\cos (\sqrt{t})}{\sqrt{t}} d t = \int \cos u 2 d u = 2 \int \cos u d u = 2 \sin u + C

u = \sqrt{t}

\int \frac{\cos (\sqrt{t})}{\sqrt{t}} d t = 2 \sin \sqrt{t} + C

karton

Expert2022-01-11Added 613 answers

$\int \frac{\cos (\sqrt{t})}{\sqrt{t}} d t \int 2 \cos (u) d u 2 \times \int \cos (u) d u 2 \sin (u) 2 \sin (\sqrt{t}) S o l u t i o n : 2 \sin (\sqrt{t}) + C$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?