[Toán 12] Tìm nguyên hàm: $\int \sin x. \sqrt{\sin^2 x+ \dfrac{1}{2}} dx.$

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truongduong9083

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$I=\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{6}} \sin x \sqrt{\sin^2x+\dfrac{1}{2}}dx$



$I=\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{6}} \sin x \sqrt{\dfrac{3}{2}-cos^2x}dx$
Đặt $ cosx=\sqrt{\dfrac{3}{2}}.sint \Longrightarrow -sinxdx=\sqrt{\dfrac{3}{2}}.costdt $
$I=-\displaystyle \dfrac{3}{2}\int^{0}_{\frac{\pi}{4}} cos^2tdt$
$I=-\displaystyle \dfrac{3}{4}\int^{0}_{\frac{\pi}{4}}(1+cos2t)dt$
$I=-\dfrac{3}{4}(t+\dfrac{sin2t}{2})|_{\frac{\pi}{4}}^0=\dfrac{3}{4}.(\dfrac{\pi}{4}+\dfrac{1}{2}) $
 
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