Find the integral of

Bernard Mora

Bernard Mora

Answered question

2022-05-06

Find the integral of 3 2 t sin ( 5 t ) with respect to t.

Answer & Explanation

Chloe Melendez

Chloe Melendez

Beginner2022-05-07Added 12 answers

Simplify.

3tsin(5t)2dt

Since 32 is constant with respect to t, move 32 out of the integral.

32tsin(5t)dt

Integrate by parts using the formula udv=uv-vdu, where u=t and dv=sin(5t).

32(t(-15cos(5t))--15cos(5t)dt)

Simplify.

32(-tcos(5t)5--cos(5t)5dt)

Since -1 is constant with respect to t, move -1 out of the integral.

32(-tcos(5t)5--cos(5t)5dt)

Simplify.

32(-tcos(5t)5+cos(5t)5dt)

Since 15 is constant with respect to t, move 15 out of the integral.

32(-tcos(5t)5+15cos(5t)dt)

Let u=5t. Then du=5dt, so 15du=dt. Rewrite using u and du.

32(-tcos(5t)5+15cos(u)15du)

Combine cos(u) and 15.

32(-tcos(5t)5+15cos(u)5du)

Since 15 is constant with respect to u, move 15 out of the integral.

32(-tcos(5t)5+15(15cos(u)du))

Simplify.

32(-tcos(5t)5+125cos(u)du)

The integral of cos(u) with respect to u is sin(u).

32(-tcos(5t)5+125(sin(u)+C))

Simplify.

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32(-cos(5t)t5+sin(u)25)+C

Replace all occurrences of u with 5t.

32(-cos(5t)t5+sin(5t)25)+C

Simplify.

-3tcos(5t)10+3sin(5t)50+C

Reorder terms.

310tcos(5t)+350sin(5t)+C

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