Find the integral: − <msqrt> − t 2 </mrow>

Sappeycuii

Sappeycuii

Answered question

2022-05-13

Find the integral: t 2 2 t d t.

Answer & Explanation

Emmy Sparks

Emmy Sparks

Beginner2022-05-14Added 17 answers

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

--t2-2tdt

Complete the square.

--(t+1)2+1dt

Let u1=t+1. Then du1=dt. Rewrite using u1 and du1.

--u12+1du1

Let u1=sin(t1), where -π2t1π2. Then du1=cos(t1)dt1. Note that since -π2t1π2cos(t1) is positive.

--sin2(t1)+1cos(t1)dt1

Simplify terms.

-cos2(t1)dt1

Use the half-angle formula to rewrite cos2(t1) as 1+cos(2t1)2.

-1+cos(2t1)2dt1

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

-(121+cos(2t1)dt1)

Split the single integral into multiple integrals.

-12(dt1+cos(2t1)dt1)

Apply the constant rule.

-12(t1+C+cos(2t1)dt1)

Let u2=2t1. Then du2=2dt1, so 12du2=dt1. Rewrite using u2 and du2.

-12(t1+C+cos(u2)12du2)

Combine cos(u2) and 12.

-12(t1+C+cos(u2)2du2)

Since 12 is constant with respect to u2, move 12 out of the integral.

-12(t1+C+12cos(u2)du2)

The integral of cos(u2) with respect to u2 is sin(u2).

-12(t1+C+12(sin(u2)+C))

Simplify.

-12(t1+12sin(u2))+C

Substitute back in for each integration substitution variable.

-12(arcsin(t+1)+12sin(2arcsin(t+1)))+C

Simplify.

-arcsin(t+1)2-sin(2arcsin(t+1))4+C

Reorder terms.

-12arcsin(t+1)-14sin(2arcsin(t+1))+C

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