Evaluate the given definite integral.∫0π2esin πx cos πx dx

An Smith

An Smith

Answered question

2022-05-16

Evaluate the given definite integral.

0π2esin πx cos πx dx

Answer & Explanation

alenahelenash

alenahelenash

Expert2022-06-05Added 556 answers

0π2esin(πx)cos(πx)dx

Let u2=sin(πx). Then du2=πcos(πx)dx, so 1πdu2=cos(πx)dx. Rewrite using u2 and d2u2.

Let u2=sin(πx). Find du2dx.

πcos(πx)

Substitute the lower limit in for x in u2=sin(πx).

ulower=sin(π0)

Simplify.

ulower=0

Substitute the upper limit in for x in u2=sin(πx).

uupper=sin(ππ2)

Simplify.

uupper=-0.97536797

The values found for ulower and uupper will be used to evaluate the definite integral.

ulower=0 uupper=-0.97536797

Rewrite the problem using u2du2, and the new limits of integration.

0-0.97536797eu21πdu2

Combine 2eu2 and 1π.

0-0.97536797eu2πdu2

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

1π0-0.97536797eu2du2

The integral of eu2 with respect to u2 is eu2.

1πeu2]0-0.97536797

Substitute and simplify.

Evaluate eu2 at -0.97536797 and at 0.

1π((e-0.97536797)-e0)

Simplify.

1π(e-0.97536797-1)

Evaluate.

Apply the distributive property.

1πe-0.97536797+1π-1

Combine1π and e-0.97536797.

e-0.97536797π+1π-1

Combine 1π and -1.

e-0.97536797π+-1π

Simplify each term.

1πe0.97536797-1π

The result can be shown in multiple forms.

Exact Form:

1πe0.97536797-1π

Decimal Form:

0.19829000

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