Find the integral: − 3 <mrow> 2 t 4

Kazeljkaml5n9y

Kazeljkaml5n9y

Answered question

2022-05-12

Find the integral: ∫ ( − 3 2 t 4 + t 2 2 + 1 2 t − sin ⁡ ( 2 t ) )  d t.

Answer & Explanation

notemilyu1208

notemilyu1208

Beginner2022-05-13Added 20 answers

Remove parentheses.

∫-32t4+t22+12t-sin(2t)dt

Split the single integral into multiple integrals.

∫-32t4dt+∫t22dt+∫12tdt+∫-sin(2t)dt

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

-∫32t4dt+∫t22dt+∫12tdt+∫-sin(2t)dt

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

-(32∫1t4dt)+∫t22dt+∫12tdt+∫-sin(2t)dt

Apply basic rules of exponents.

-32∫t-4dt+∫t22dt+∫12tdt+∫-sin(2t)dt

By the Power Rule, the integral of t-4 with respect to t is -13t-3.

-32(-13t-3+C)+∫t22dt+∫12tdt+∫-sin(2t)dt

Since 12 is constant with respect to t, move 12 out of the integral.

-32(-13t-3+C)+12∫t2dt+∫12tdt+∫-sin(2t)dt

By the Power Rule, the integral of t2 with respect to t is 13t3.

-32(-13t-3+C)+12(13t3+C)+∫12tdt+∫-sin(2t)dt

Since 12 is constant with respect to t, move 12 out of the integral.

-32(-13t-3+C)+12(13t3+C)+12∫1tdt+∫-sin(2t)dt

The integral of 1t with respect to t is ln(|t|).

-32(-13t-3+C)+12(13t3+C)+12(ln(|t|)+C)+∫-sin(2t)dt

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

-32(-13t-3+C)+12(13t3+C)+12(ln(|t|)+C)-∫sin(2t)dt

Let u=2t. Then du=2dt, so 12du=dt. Rewrite using u and du.

-32(-13t-3+C)+12(13t3+C)+12(ln(|t|)+C)-∫sin(u)12du

Combine sin(u) and 12.

-32(-13t-3+C)+12(13t3+C)+12(ln(|t|)+C)-∫sin(u)2du

Since 12 is constant with respect to u, move 12 out of the integral.

-32(-13t-3+C)+12(13t3+C)+12(ln(|t|)+C)-(12∫sin(u)du)

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

-32(-13t-3+C)+12(13t3+C)+12(ln(|t|)+C)-12(-cos(u)+C)

Simplify.

12t3+t36+ln(|t|)2+cos(u)2+C

Replace all occurrences of u with 2t.

12t3+t36+ln(|t|)2+cos(2t)2+C

Reorder terms.

12t3+16t3+12ln(|t|)+12cos(2t)+C

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