Answer to Question #224864 in Chemical Engineering for Lokika

Question

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

Answer 1

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.

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