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kwisangqaquqw3

kwisangqaquqw3

Answered question

2022-05-12

Evaluate 5 9 cos 1 ( t ) d t.

Answer & Explanation

Jerry Kidd

Jerry Kidd

Beginner2022-05-13Added 18 answers

Since 59 is constant with respect to t, move 59 out of the integral.

59cos-1(t)dt

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

59(cos-1(t)t-t(-11-t2)dt)

Combine t and 11-t2.

59(cos-1(t)t--t1-t2dt)

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

59(cos-1(t)t--t1-t2dt)

Simplify.

59(cos-1(t)t+t1-t2dt)

Let u=1-t2. Then du=-2tdt, so -12du=tdt. Rewrite using u and du.

59(cos-1(t)t+1u1-2du)

Simplify.

59(cos-1(t)t+-12udu)

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

59(cos-1(t)t-12udu)

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

59(cos-1(t)t-(121udu))

Apply basic rules of exponents.

59(cos-1(t)t-12u-12du)

By the Power Rule, the integral of u-12 with respect to u is 2u12.

59(cos-1(t)t-12(2u12+C))

Rewrite 59(cos-1(t)t-12(2u12+C)) as 59(cos-1(t)t-u12)+C.

59(cos-1(t)t-u12)+C

Replace all occurrences of u with 1-t2.

59(cos-1(t)t-(1-t2)12)+C

Simplify.

5(tcos-1(t)-(1-t2)12)9+C

Reorder terms.

59(tcos-1(t)-(1-t2)12)+C

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