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Daphne Fry 2022-05-11 Answered
Evaluate 5 2 cos ( 4 x ) ln ( sin ( 4 x ) + 1 ) d x.
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Answers (1)

Emmy Sparks
Answered 2022-05-12 Author has 18 answers

Simplify.

5cos(4x)ln(sin(4x)+1)2dx

Since 52 is constant with respect to x, move 52 out of the integral.

52cos(4x)ln(sin(4x)+1)dx

Integrate by parts using the formula udv=uv-vdu, where u=ln(sin(4x)+1) and dv=cos(4x).

52(ln(sin(4x)+1)(14sin(4x))-14sin(4x)4cos(4x)1+sin(4x)dx)

Simplify.

52(ln(sin(4x)+1)sin(4x)4-sin(4x)cos(4x)1+sin(4x)dx)

Let u2=sin(4x). Then du2=4cos(4x)dx14du2=cos(4x)dx. Rewrite using u2 and du2.

52(ln(sin(4x)+1)sin(4x)4-u21+u214du2)

52(ln(sin(4x)+1)sin(4x)4-u24(1+u2)du2)

Since 14 is constant with respect to u2, move 14 out of the integral.

52(ln(sin(4x)+1)sin(4x)4-(14u21+u2du2))

Reorder 1 and u2.

52(ln(sin(4x)+1)sin(4x)4-14u2u2+1du2)

Divide u2 by u2+1.

52(ln(sin(4x)+1)sin(4x)4-141-1u2+1du2)

Split the single integral into multiple integrals.

52(ln(sin(4x)+1)sin(4x)4-14(du2+-1u2+1du2))

Apply the constant rule.

52(ln(sin(4x)+1)sin(4x)4-14(u2+C+-1u2+1du2))

Since -1 is constant with respect to u2, move -1 out of the integral.

52(ln(sin(4x)+1)sin(4x)4-14(u2+C-1u2+1du2))

Let u3=u2+1. Then du3=du2. Rewrite using u3 and du3.

52(ln(sin(4x)+1)sin(4x)4-14(u2+C-1u3du3))

The integral of 31u3 with respect to u3 is ln(|u3|).

52(ln(sin(4x)+1)sin(4x)4-14(u2+C-(ln(|u3|)+C)))

Simplify.

52(ln(sin(4x)+1)sin(4x)4-u24+ln(|u3|)4)+C

Substitute back in for each integration substitution variable.

52(ln(sin(4x)+1)sin(4x)4-sin(4x)4+ln(|sin(4x)+1|)4)+C

Simplify.

5(sin(4x)ln(sin(4x)+1)-sin(4x)+ln(|sin(4x)+1|))8+C

Reorder terms.

58(sin(4x)ln(sin(4x)+1)-sin(4x)+ln(|sin(4x)+1|))+C

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