Question

Evaluate the following integral :

$\sf\int \dfrac{\sin x}{\sqrt{1+\sin x}}$
​

Answer 1

Answer:

hope it help you dear please

Evaluate the following integral :

\sf\int \dfrac{\sin x}{\sqrt{1+\sin x}}∫1+sinxsinx

Answer 2

fistly

$Let \bf{I_1}=\int \sqrt{1+\sin x} \;dx and \bf{I_2}=\int \dfrac{1 }{\sqrt{1+\sin x}}dx</p><p></p><p>Solving I₁ :</p><p></p><p>\bf{I_1}=\int \sqrt{1+\sin x} \;dx</p><p></p><p>{ I_1} = \sqrt{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2} + 2\sin^2\dfrac{x}{2}\cos^2\dfrac{x}{2}} dx</p><p></p><p>{ I_1} = \sqrt{\left(\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}\right)^2} dx</p><p></p><p>I_1 = \int\left( \sin \dfrac{x}{2} +\cos\dfrac{x}{2}\right)dx</p><p></p><p>I_1 = -\dfrac{\cos \dfrac{x}{2}}{\dfrac{1}{2}}+\dfrac{\sin \dfrac{x}{2}}{\dfrac{1}{2}}</p><p></p><p>I_1 = -2\cos\dfrac{x}{2}+2\sin \dfrac{x}{2}</p><p></p><p>I_1 =2\left[sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right]</p><p></p><p>Now, Solving for I₂ :</p><p></p><p>\bf{I_2}=\int \dfrac{1 }{\sqrt{1+\sin x}}dx</p><p></p><p>{I_2}=\int \dfrac{1 }{\sin\dfrac{x}{2}+\cos\dfrac{x}{2}}dx {As solved in I₁}</p><p></p><p>Multiplying and Dividing by 1/√2</p><p></p><p>{I_2}=\dfrac{1}{\sqrt2}\int \dfrac{1 }{\sin\dfrac{x}{2}.\dfrac{1}{\sqrt2}+\cos\dfrac{x}{2}.\dfrac{1}{\sqrt2}}dx </p><p></p><p>{I_2}=\dfrac{1}{\sqrt2}\int \dfrac{1 }{\sin\dfrac{x}{2}.\sin\dfrac{\pi}{4}+\cos\dfrac{x}{2}.\cos\dfrac{\pi}{4}}dx</p><p></p><p>I_2 =\dfrac{1}{\sqrt2}\int\dfrac{1}{\cos\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)} dx</p><p></p><p>I_2 = \dfrac{1}{\sqrt2}\int \sec\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right) dx</p><p></p><p>I_2 = \dfrac{1}{\sqrt2}\dfrac{\log\left|\sec\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)+\tan\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)\right|}{\dfrac{1}{2}}</p><p></p><p>I_2= \sqrt2\log\left|\sec\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right) +\tan\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)\right|.</p><p></p><p>Putting I₁ and I₂ in (1) :</p><p></p><p>\bf{I}=\int \sqrt{1+\sin x} \;dx-\int \dfrac{1 }{\sqrt{1+\sin x}}dx \quad\cdots(i)</p><p></p><p>I = I₁ + I₂</p><p></p><p>I =2\left[sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right] - \sqrt2\log\left|\sec\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right) +\tan\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)\right| + c</p><p></p><p>Therefore,</p><p>\pink{\int \dfrac{\sin x }{\sqrt{1+\sin x}} dx }=\green{2\left[sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right] - \sqrt2\log\left|\sec\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right) +\tan\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)\right| +c}</p><p></p><p>$

11 Pm Ho gye aa

It's The last hour of ur birthday 2022 ❤️

Last time ... I Wish u a very happiest birthday ❣️

But...,

Mai ki kita eho ja ki tu c mainu eni vadi saza de rhe oo ಥ_ಥ

Please onn Aajo

Mera dil ni laghda tuhade bina (ᗒ-ᗕ)

:'( ❣️