∫3−x3+xsin−1163−xdx

I=∫3−x3+xdx Put x=3cos⁡2θ⇒θ=12cos−1⁡x3dx=−6sin⁡2θdθI=∫3−3cos⁡2θ3+3cos⁡2θ(−6sin⁡2θ)dθ==−6∫1−cos⁡2θ1+cos⁡θsin⁡2θdθ=−6∫2sin2⁡θ2cos2⁡θsin⁡2θdθ=−6∫sin⁡θcos⁡θ×2sin⁡θdθ=−12∫sin2⁡θdθ=−12∫1−cos⁡2θ2dθ=−6∫1dθ−∫cos⁡2θdθ=−6θ−sin⁡2θ2+C=−612cos−1⁡x3−121−x29+C=−3cos−1⁡x3+9−x2+C

∫3−x3+xsin−1163−xdx

I=∫3−x3+xdx Put x=3cos⁡2θ⇒θ=12cos−1⁡x3dx=−6sin⁡2θdθI=∫3−3cos⁡2θ3+3cos⁡2θ(−6sin⁡2θ)dθ==−6∫1−cos⁡2θ1+cos⁡θsin⁡2θdθ=−6∫2sin2⁡θ2cos2⁡θsin⁡2θdθ=−6∫sin⁡θcos⁡θ×2sin⁡θdθ=−12∫sin2⁡θdθ=−12∫1−cos⁡2θ2dθ=−6∫1dθ−∫cos⁡2θdθ=−6θ−sin⁡2θ2+C=−612cos−1⁡x3−121−x29+C=−3cos−1⁡x3+9−x2+C

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$\int \sqrt{\frac{3 - x}{3 + x}} \sin^{- 1} (\frac{1}{\sqrt{6}} \sqrt{3 - x}) d x$

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$- 3 \cos^{- 1} (\frac{x}{3}) + \sqrt{9 - x^{2}} + C$

$- 6 \cos^{- 1} (\frac{x}{3}) + 6 \sqrt{9 - x^{2}} + C$

$2 \cos^{- 1} (\frac{x}{6}) + 3 \sqrt{9 - x^{2}} + C$

$3 \cos^{- 1} \frac{x}{3} - \sqrt{9 - x^{2}} + C$

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Detailed Solution

$I = \int \sqrt{\frac{3 - x}{3 + x}} d x$

$\begin{array}{l} Put x = 3 \cos 2 θ \Rightarrow θ = \frac{1}{2} \cos^{- 1} (\frac{x}{3}) \\ d x = - 6 \sin 2 θ d θ \\ I = \int \sqrt{\frac{3 - 3 \cos 2 θ}{3 + 3 \cos 2 θ}} (- 6 \sin 2 θ) d θ \\ == - 6 \int \sqrt{\frac{1 - \cos 2 θ}{1 + \cos θ}} \sin 2 θ d θ \\ = - 6 \int \sqrt{\frac{2 \sin^{2} θ}{2 \cos^{2} θ}} \sin 2 θ d θ \\ = - 6 \int \frac{\sin θ}{\cos θ} \times 2 \sin θ d θ \\ = - 12 \int \sin^{2} θ d θ \end{array}$

$\begin{array}{l} = - 12 \int [\frac{1 - \cos 2 θ}{2}] d θ \\ = - 6 [\int 1 d θ - \int \cos 2 θ d θ] \\ = - 6 [θ - \frac{\sin 2 θ}{2}] + C \\ = - 6 [\frac{1}{2} \cos^{- 1} \frac{x}{3} - \frac{1}{2} \sqrt{1 - \frac{x^{2}}{9}}] + C \\ = [- 3 \cos^{- 1} \frac{x}{3} + \sqrt{9 - x^{2}}] + C \end{array}$