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Question Number 123133 by Khalmohmmad last updated on 23/Nov/20

∫(√(sin x))dx=?

$$\int\sqrt{\mathrm{sin}\:{x}}{dx}=? \\ $$

Commented by MJS_new last updated on 23/Nov/20

see question 84607

$$\mathrm{see}\:\mathrm{question}\:\mathrm{84607} \\ $$

Commented by Dwaipayan Shikari last updated on 23/Nov/20

(√(sinx))=(√(1−msin^2 θ))  1−sinx=msin^2 θ  (sin(x/2)−cos(x/2))^2 =msin^2 θ  2(sin^2 ((x/2)−(π/4)))=msin^2 θ       θ=((x/2)−(π/4))   m=2  E(z∣m)=∫_0 ^z (√(1−msin^2 x)) dx  ∫(√(sinx)) dx=2E((x/2)−(π/4)∣2)

$$\sqrt{{sinx}}=\sqrt{\mathrm{1}−{msin}^{\mathrm{2}} \theta} \\ $$$$\mathrm{1}−{sinx}={msin}^{\mathrm{2}} \theta \\ $$$$\left({sin}\frac{{x}}{\mathrm{2}}−{cos}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} ={msin}^{\mathrm{2}} \theta \\ $$$$\mathrm{2}\left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\right)={msin}^{\mathrm{2}} \theta\:\:\:\:\: \\ $$$$\theta=\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\:\:\:{m}=\mathrm{2} \\ $$$${E}\left({z}\mid{m}\right)=\int_{\mathrm{0}} ^{{z}} \sqrt{\mathrm{1}−{msin}^{\mathrm{2}} {x}}\:{dx} \\ $$$$\int\sqrt{{sinx}}\:{dx}=\mathrm{2}{E}\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right) \\ $$

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