Mathematics

$$\int_0^8 {\frac{{\sqrt x }}{{\sqrt x  + \sqrt {8 - x} }}dx} $$


ANSWER

4


SOLUTION

$$ \int_{0}^{8}  \dfrac{ \sqrt{x}}{ \sqrt{x} + \sqrt{8-x}} dx $$

$$ \int_{0}^{8}  \dfrac{ \sqrt{x}}{ \sqrt{x} + \sqrt{8-x}}  \times  \dfrac{ ( \sqrt{x} - \sqrt{8-x}) }{ ( \sqrt{x} - \sqrt{8-x})} dx  $$

$$ \int_{0}^{8}  \dfrac{ x - \sqrt{8x – x^2}}{ 2x-8} dx $$

$$ \dfrac{1}{2}  \int_{0}^{8}  \dfrac{ x - \sqrt{8x – x^2}}{ x-4} dx $$

Put $$x – 4 = u =   \implies   dx =  du$$ and  $$x^ 2 =  (u+4)^2 $$

$$ = \dfrac{1}{2}  \dfrac{ u + 4 - \sqrt{ 8u + 32 – (u+4)^2}}{u} du $$

$$ = \dfrac{1}{2} \int du +  2 \int \dfrac{du}{u} - \dfrac{1}{2} \int  \dfrac{ \sqrt{ - (u+4)^2 + 8u + 32}}{u} du $$

$$v = u^2 \implies  du = \dfrac{1}{2u} dv  $$ and $$ \dfrac{1}{ u^2} = \dfrac{1}{v}$$

$$ = \dfrac{u}{2} + 2 log u - \dfrac{1}{4} \int \dfrac{ \sqrt{16-v}}{v} dv $$

$$=\ dfrac{u}{2} + 2 log u - \dfrac{1}{2} \int \dfrac{ w^2}{w^2-16} dw $$      put  $$ \sqrt{ 16 – v} = w$$   and $$dv =  -2 \sqrt{16}-v dw$$

$$ = \dfrac{u}{2} + 2 log u - \dfrac{1}{2} \int \dfrac{ w^2+16-16}{w^2-16} dw $$

$$= \dfrac{u}{2} + 2 log u - \dfrac{1}{2} w – 8 \int \dfrac{dw}{w^2 -16} $$

$$= \dfrac{u}{2} + 2 log u - \dfrac{1}{2}  \sqrt{ 16 – u^2} – 8 \times \dfrac{1}{2} log | \dfrac{w-4}{w+4}|$$

$$ = \dfrac{x-4}{2} + 2 log |x-4| - \dfrac{1}{2} \sqrt{ 16 – (x-4)^2} – log | \dfrac{ \sqrt{16 – u^2}-4}{ \sqrt{16-u^2} + 4} |$$

$$ = \dfrac{x-4}{2} + 2 log |x-4| - \dfrac{1}{2} \sqrt{ 8x – x^2} – log | \dfrac{ \sqrt{8x – x^2}-4}{ \sqrt{8x-x^2} + 4} |_{0}^{8}$$

$$ = \dfrac{8-4}{2} + 2 log |8-4| - \dfrac{1}{2} \sqrt{ 8.8 – (8)^2} – log | \dfrac{ \sqrt{8.8 – 8^2}-4}{ \sqrt{8.8-8^2} + 4} |$$

$$ - ( -2  + 2 log 4 -0) = 4$$

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