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

$$\underset{0}{\stackrel{\pi }{\int }}\frac{\sin {}^{2}x}{\sqrt{x}}dx$$

Answered by mindispower last updated on 29/Nov/20

$${sin}^2\left(x\right)=\frac{1-cos\left(2x\right)}{2}$$$$\iff {\int }_0^{\pi }\frac{1-cos\left(2x\right)}{2\sqrt{x}}dx$$$$={\int }_0^{\pi }\frac{1}{2\sqrt{x}}-{\int }_0^{\pi }\frac{cos\left(2x\right)}{2\sqrt{x}}dx=A-B$$$$x=\frac{\pi t^2}{4}$$$$={\left[\sqrt{x}\right]}_0^{\pi }-{\int }_0^2\frac{cos\left(\frac{\pi }{2}{t}^2\right)}{\sqrt{\pi }t}.\frac{\pi }{2}tdt$$$$=\sqrt{\pi }-{\int }_0^2\frac{cos\left(\frac{\pi t^2}{2}\right)}{2}\sqrt{\pi }dt$$$$=\sqrt{\pi }-\frac{\sqrt{\pi }}{2}{\int }_0^{2}cos\left(\pi \frac{t^2}{2}\right)dt$$$$Recall{\int }_0^{u}cos\left(\pi \frac{t^2}{2}\right)dt=C\left(u\right)Ferneslintegral$$$$weget$$$${\int }_0^{\pi }\frac{{sin}^2\left(t\right)}{2\sqrt{t}}dt=\sqrt{\pi }-\frac{\sqrt{\pi }}{2}C\left(2\right)=\frac{\sqrt{\pi }}{2}(2-C\left(2\right))$$

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