Mathematics

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

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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One Word Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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