Question #224864

Evaluate ∫0^π\2α α sin αx dx,α>0 by using Leibniz formula.

Expert's answer

Compute the definite integral:

"\\int_0^{\\frac{1}{2}\\pi \\alpha} \\alpha \\sin (\\alpha x)dx\\\\"

Factor out constants:

"\\alpha \\int_0^{\\frac{1}{2}\\pi \\alpha} \\sin (\\alpha x)dx\\\\"

For the integrand "\\sin \\alpha x" , substitute "u= \\alpha x" and "du = \\alpha dx" . This gives a new lower bound "u=0 \\space \\alpha = 0" and upper bound "u= \\alpha (0.5 \\pi \\alpha)= 0.5 \\pi \\alpha^2"

"\\int_0^{\\frac{1}{2}\\pi \\alpha^2} \\sin (u)du\\\\"

Apply the fundamental theorem of calculus.

The antiderivative of "sin(u) \\space is -cos(u)"

"=(-\\cos (u))|_{0}^{0.5 \\pi \\alpha^2}"

Evaluate the antiderivative at the limits and subtract.

"=(-\\cos (u))|_{0}^{0.5 \\pi \\alpha^2}= (-\\cos (0.5 \\pi \\alpha^2))-(-\\cos (0))\\\\\n=1-\\cos(0.5 \\pi \\alpha^2)"

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