Evaluate − 3 <msqrt> − 2 x 2 </mrow>

sembuang711q6

sembuang711q6

Answered question

2022-05-11

Evaluate 3 2 x 2 + 4 x + 5 d x.

Answer & Explanation

Percyaehyq

Percyaehyq

Beginner2022-05-12Added 18 answers

Since -3 is constant with respect to x, move -3 out of the integral.

-3-2x2+4x+5dx

Complete the square.

-3-2(x-1)2+7dx

Let u1=x-1. Then du1=dx. Rewrite using u1 and du1.

-3-2u12+7du1

Let u1=72sin(t), where -π2tπ2. Then du1=14cos(t)2dt. Note that since -π2tπ214cos(t)2 is positive.

-3-2(72sin(t))2+7214cos(t)2dt

Simplify terms.

-398cos2(t)2dt

Since 982 is constant with respect to t, move 982 out of the integral.

-3(982cos2(t)dt)

Simplify.

-3982cos2(t)dt

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

-39821+cos(2t)2dt

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

-3982(121+cos(2t)dt)

Simplify.

-39841+cos(2t)dt

Split the single integral into multiple integrals.

-3984(dt+cos(2t)dt)

Apply the constant rule.

-3984(t+C+cos(2t)dt)

Let u2=2t. Then du2=2dt, so 12du2=dt. Rewrite using u2 and du2.

-3984(t+C+cos(u2)12du2)

Combine cos(u2) and 12.

-3984(t+C+cos(u2)2du2)

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

-3984(t+C+12cos(u2)du2)

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

-3984(t+C+12(sin(u2)+C))

Simplify.

-2124(t+12sin(u2))+C

Substitute back in for each integration substitution variable.

-2124(arcsin(2(x-1)7)+12sin(2arcsin(2(x-1)7)))+C

Simplify.

-2124(arcsin(147(x-1))+12sin(2arcsin(147(x-1))))+C

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