myntfalskj4

Answered

2022-07-02

How does ${\int }_{\pi /3}^{\pi /2}\frac{1-{\mathrm{cos}}^{2}x}{\sqrt{{\mathrm{sin}}^{2}\left(x/2\right)}}dx$ simplify to ${\int }_{\pi /3}^{\pi /2}4\mathrm{sin}\left(x/2\right){\mathrm{cos}}^{2}\left(x/2\right)dx$?

Answer & Explanation

Kiley Hunter

Expert

2022-07-03Added 7 answers

$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}4\mathrm{sin}\left(x/2\right){\mathrm{cos}}^{2}\left(x/2\right)dx={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{4{\mathrm{sin}}^{2}\left(x/2\right){\mathrm{cos}}^{2}\left(x/2\right)}{\mathrm{sin}\left(x/2\right)}dx$
$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{{\left[2\mathrm{sin}\left(x/2\right)\mathrm{cos}\left(x/2\right)\right]}^{2}}{\mathrm{sin}\left(x/2\right)}dx$
$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}\left(x\right)}{\mathrm{sin}\left(x/2\right)}dx$
$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1-{\mathrm{cos}}^{2}\left(x\right)}{\sqrt{{\mathrm{sin}}^{2}\left(x/2\right)}}dx$
I used $2\mathrm{sin}\left(x/2\right)\mathrm{cos}\left(x/2\right)=\mathrm{sin}\left(x\right)$ and $1={\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)$

prirodnogbk

Expert

2022-07-04Added 6 answers

$\frac{1-{\mathrm{cos}}^{2}x}{\mathrm{sin}\frac{x}{2}}=\frac{{\mathrm{sin}}^{2}x}{\mathrm{sin}\frac{x}{2}}=\frac{{\left(2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}\right)}^{2}}{\mathrm{sin}\frac{x}{2}}=4{\mathrm{cos}}^{2}\frac{x}{2}\mathrm{sin}\frac{x}{2}.$

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