Evaluate the integral.∫0In 22e-xcosh x dx

An Smith

An Smith

Answered question

2022-05-16

Evaluate the integral.

0In 22e-xcosh x dx

Answer & Explanation

alenahelenash

alenahelenash

Expert2022-06-05Added 556 answers

0ln(2)2e-xcos(x)dx

Since 2 is constant with respect to x, move 2 out of the integral.

20ln(2)e-xcos(x)dx

Reorder e-x and cos(x).

20ln(2)cos(x)e-xdx

Integrate by parts using the formula udv=uv-vdu, where u=cos(x) and dv=e-x.

2(cos(x)(-e-x)]0ln(2)-0ln(2)-e-x(-sin(x))dx)

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

2(cos(x)(-e-x)]0ln(2)-(--10ln(2)e-x(sin(x))dx))

Simplify the expression.

2(cos(x)(-e-x)]0ln(2)-0ln(2)sin(x)e-xdx)

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

2(cos(x)(-e-x)]0ln(2)-(sin(x)(-e-x)]0ln(2)-0ln(2)-e-xcos(x)dx))

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

2(cos(x)(-e-x)]0ln(2)-(sin(x)(-e-x)]0ln(2)--0ln(2)e-xcos(x)dx))

Simplify by multiplying through.

2(cos(x)(-e-x)]0ln(2)-(sin(x)(-e-x)]0ln(2))-0ln(2)e-xcos(x)dx)

Solving for 0ln(2)e-xcos(x)dx, we find that 0ln(2)e-xcos(x)dx cos(x)(-e-x)]0ln(2)-(sin(x)(-e-x)]0ln(2))2.

2(cos(x)(-e-x)]0ln(2)-(sin(x)(-e-x)]0ln(2))2]0ln(2))

Substitute and simplify.

cos(ln(2))(-e-ln(2))+cos(0)-(sin(ln(2))(-e-ln(2))+sin(0))-(cos(ln(2))(-e-ln(2))+cos(0)-(sin(ln(2))(-e-ln(2))+sin(0)))

Simplify.

cos(ln(2))(-e-ln(2))+1+sin(ln(2))e-ln(2)-(cos(ln(2))(-e-ln(2))+1+sin(ln(2))e-ln(2))

Simplify.

= 0 - Answer

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