find-0-pi-sinx-1-sin-2-x-dx-

Question Number 29443 by prof Abdo imad last updated on 08/Feb/18

f i n d \int_{0}^{π} \frac{s i n x}{\sqrt{1 + {s i n}^{2} x}} d x

Answered by sma3l2996 last updated on 09/Feb/18

t = c o s x \Rightarrow d t = - s i n x d x

\int_{0}^{π} \frac{s i n x}{\sqrt{1 + {s i n}^{2} x}} d x = \int_{- 1}^{1} \frac{d t}{\sqrt{1 + (1 - t^{2})}} = \int_{- 1}^{1} \frac{d t}{\sqrt{2} \sqrt{1 - {(\frac{t}{\sqrt{2}})}^{2}}}

t = \sqrt{2} u \Rightarrow d t = \sqrt{2} d u

\int_{0}^{π} \frac{s i n x}{\sqrt{1 + {s i n}^{2} x}} d x = \int_{- \sqrt{2} / 2}^{\sqrt{2} / 2} \frac{d u}{\sqrt{1 - u^{2}}}

u = s i n y \Rightarrow d y = \frac{d u}{\sqrt{1 - u^{2}}}

\int_{0}^{π} \frac{s i n x}{\sqrt{1 + {s i n}^{2} x}} d x = \int_{- π / 4}^{π / 4} d y = \frac{π}{2}