Compute 7 6 &#xFEFF; </mrow> t cos − 1 </

encamineu2cki

encamineu2cki

Answered question

2022-05-13

Compute 7 6  t cos 1 ( t ) d t.

Answer & Explanation

Brennan Frye

Brennan Frye

Beginner2022-05-14Added 20 answers

Simplify.

7tcos-1(t)6dt

Since 76 is constant with respect to t, move 76 out of the integral.

76tcos-1(t)dt

Integrate by parts using the formula udv=uv-vdu, where u=cos-1(t) and dv=t.

76(cos-1(t)(12t2)-12t2(-11-t2)dt)

Simplify.

76(cos-1(t)t22--t221-t2dt)

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

76(cos-1(t)t22--t221-t2dt)

Simplify.

76(cos-1(t)t22+t221-t2dt)

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

76(cos-1(t)t22+12t21-t2dt)

Let t=sin(t1), where -π2t1π2. Then dt=cos(t1)dt1. Note that since -π2t1π2cos(t1) is positive.

76(cos-1(t)t22+12sin2(t1)1-sin2(t1)cos(t1)dt1)

Simplify terms.

76(cos-1(t)t22+12sin2(t1)dt1)

Use the half-angle formula to rewrite sin2(t1) as 1-cos(2t1)2.

76(cos-1(t)t22+121-cos(2t1)2dt1)

Since 12 is constant with respect to t1, move 12 out of the integral.

76(cos-1(t)t22+12(121-cos(2t1)dt1))

Simplify.

76(cos-1(t)t22+141-cos(2t1)dt1)

Split the single integral into multiple integrals.

76(cos-1(t)t22+14(dt1+-cos(2t1)dt1))

Apply the constant rule.

76(cos-1(t)t22+14(t1+C+-cos(2t1)dt1))

Since -1 is constant with respect to t1, move -1 out of the integral.

76(cos-1(t)t22+14(t1+C-cos(2t1)dt1))

Let u=2t1. Then du=2dt1, so 12du=dt1. Rewrite using u and du.

76(cos-1(t)t22+14(t1+C-cos(u)12du))

Combine cos(u) and 12.

76(cos-1(t)t22+14(t1+C-cos(u)2du))

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

76(cos-1(t)t22+14(t1+C-(12cos(u)du)))

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

76(cos-1(t)t22+14(t1+C-12(sin(u)+C)))

Simplify.

76(cos-1(t)t22+t14-sin(u)8)+C

Substitute back in for each integration substitution variable.

76(cos-1(t)t22+sin-1(t)4-sin(2sin-1(t))8)+C

Simplify.

76(12cos-1(t)t2+14sin-1(t)-18sin(2sin-1(t)))+C

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