Compute 3 2 &#xFEFF; </mrow> sin − 1 </mrow

Paul Duran

Paul Duran

Answered question

2022-05-12

Compute 3 2  sin 1 ( t ) d t.

Answer & Explanation

Finnegan Zimmerman

Finnegan Zimmerman

Beginner2022-05-13Added 16 answers

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

32sin-1(t)dt

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

32(sin-1(t)t-t11-t2dt)

Combine t and 11-t2.

32(sin-1(t)t-t1-t2dt)

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

32(sin-1(t)t-1u1-2du)

Simplify.

32(sin-1(t)t--12udu)

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

32(sin-1(t)t--12udu)

Simplify.

32(sin-1(t)t+12udu)

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

32(sin-1(t)t+121udu)

Apply basic rules of exponents.

32(sin-1(t)t+12u-12du)

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

32(sin-1(t)t+12(2u12+C))

Rewrite 32(sin-1(t)t+12(2u12+C)) as 32(sin-1(t)t+u12)+C.

32(sin-1(t)t+u12)+C

Replace all occurrences of u with 1-t2.

32(sin-1(t)t+(1-t2)12)+C

Simplify.

3(tsin-1(t)+(1-t2)12)2+C

Reorder terms.

32(tsin-1(t)+(1-t2)12)+C

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?