0-pi-sin-2-x-x-dx-

Question Number 123896 by john_santu last updated on 29/Nov/20

\underset{0}{\int^{π}} \frac{\sin^{2} x}{\sqrt{x}} d x

Answered by mindispower last updated on 29/Nov/20

{s i n}^{2} (x) = \frac{1 - c o s (2 x)}{2}

\Leftrightarrow \int_{0}^{π} \frac{1 - c o s (2 x)}{2 \sqrt{x}} d x

= \int_{0}^{π} \frac{1}{2 \sqrt{x}} - \int_{0}^{π} \frac{c o s (2 x)}{2 \sqrt{x}} d x = A - B

x = \frac{π t^{2}}{4}

= {[\sqrt{x}]}_{0}^{π} - \int_{0}^{2} \frac{c o s (\frac{π}{2} t^{2})}{\sqrt{π} t} . \frac{π}{2} t d t

= \sqrt{π} - \int_{0}^{2} \frac{c o s (\frac{π t^{2}}{2})}{2} \sqrt{π} d t

= \sqrt{π} - \frac{\sqrt{π}}{2} \int_{0}^{2} c o s (π \frac{t^{2}}{2}) d t

R e c a l l \int_{0}^{u} c o s (π \frac{t^{2}}{2}) d t = C (u) F e r n e s l i n t e g r a l

w e g e t

\int_{0}^{π} \frac{{s i n}^{2} (t)}{2 \sqrt{t}} d t = \sqrt{π} - \frac{\sqrt{π}}{2} C (2) = \frac{\sqrt{π}}{2} (2 - C (2))