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Question Number 107069 by mathdave last updated on 08/Aug/20

Answered by bemath last updated on 09/Aug/20

$$@bemath@$$$${(\sqrt{\cos x}+\sqrt{\sin x})}^5=(\sqrt{\cos x}{(1+\sqrt{\tan x})}^5$$$$=\cos {}^{2}x\sqrt{\cos x}{(1+\sqrt{\tan x})}^5$$$$I=\int \frac{\sqrt{\tan x}\sec {}^{2}xdx}{{(1+\sqrt{\tan x})}^5}$$$$set1+\sqrt{\tan x}=z\Rightarrow \sqrt{\tan x}=z-1$$$$\frac{\sec {}^{2}xdx}{2\sqrt{\tan x}}=dz\Rightarrow dx=\frac{2\sqrt{\tan x}dz}{\sec {}^{2}x}$$$$I=\int \frac{\sqrt{\tan x}\sec {}^{2}x}{z^5}.\left(\frac{2\sqrt{\tan x}}{\sec {}^{2}x}dz\right)$$$$I=\int \frac{2{(z-1)}^{2}dz}{z^5}$$$$I=2\int \frac{z^2-2z+1}{z^5}dz=2\int z^{-3}-2z^{-4}+z^{-5}dz$$$$I=2(-\frac{1}{2z^2}+\frac{2}{3z^3}-\frac{1}{4z^4})+C$$$$I=-\frac{1}{z^2}+\frac{4}{3z^3}-\frac{1}{2z^4}+C$$$$I=\frac{-6z^2+8z-3}{6z^4}+C$$$$I=\frac{-6{(1+\sqrt{\tan x})}^2+8\sqrt{\tan x}+5}{6{(1+\sqrt{\tan x})}^4}+C$$

Commented by bobhans last updated on 09/Aug/20

$$\mathrm{I}={\left[\frac{-6{(1+\sqrt{\tan \mathrm{x}})}^2+8\sqrt{\tan \mathrm{x}}+5}{6{(1+\sqrt{\tan \mathrm{x}})}^4}\right]}_0^{\frac{\pi }{4}}$$$$\mathrm{I}=\left[\frac{-6\left(4\right)+8+5}{6.16}\right]-\left[\frac{-6+5}{6}\right]$$$$\mathrm{I}=-\frac{11}{96}+\frac{1}{6}=\frac{-11+16}{96}=\frac{5}{96}$$

Answered by Tony6400 last updated on 08/Aug/20

$$\mathrm{Evaluate}{\int }_0^{\frac{\pi }{4}}\frac{\sqrt{\mathrm{sinx}}\mathrm{dx}}{{(\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}})}^5}$$$$$$$$\mathrm{Take}{(\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}})}^5={\left[\sqrt{\mathrm{cosx}}(1+\frac{\sqrt{\mathrm{sinx}}}{\sqrt{\mathrm{cosx}}})\right]}^5={\left[\sqrt{\mathrm{cosx}}\right]}^4.\sqrt{\mathrm{cosx}}{[1+\sqrt{\mathrm{tanx}}]}^5$$$$={\cos }^2\mathrm{x}\sqrt{\mathrm{cosx}}{(1+\sqrt{\mathrm{tanx}})}^5$$$$\therefore I={\int }_0^{\frac{\pi }{4}}\frac{\sqrt{\mathrm{sinx}}}{{\cos }^2\mathrm{x}\sqrt{\mathrm{cosx}}{(1+\sqrt{\mathrm{tanx}})}^5}\mathrm{dx}$$$$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\frac{\sqrt{\mathrm{tanx}}.{\sec }^2\mathrm{x}}{{(1+\sqrt{\mathrm{tanx}})}^5}\mathrm{dx}$$$$\mathrm{Let}\mathrm{w}=1+\sqrt{\mathrm{tanx}}\Rightarrow \frac{\mathrm{dw}}{\mathrm{dx}}=\frac{{\sec }^2\mathrm{x}}{2\sqrt{\mathrm{tanx}}}\Rightarrow \mathrm{dx}=\frac{2\sqrt{\mathrm{tanx}}}{{\sec }^2\mathrm{x}}\mathrm{dw}$$$$\mathrm{When}\mathrm{x}=\frac{\pi }{4},\mathrm{w}=2.\mathrm{When}$$$$\mathrm{x}=0,\mathrm{w}=1\Rightarrow I={\int }_1^2\frac{\sqrt{\mathrm{tanx}}.{\sec }^2\mathrm{x}}{{\mathrm{w}}^5}.\frac{2\sqrt{\mathrm{tanx}}}{{\sec }^2\mathrm{x}}\mathrm{dw}$$$$I=2{\int }_1^2\frac{\mathrm{tanx}}{{\mathrm{w}}^5}\mathrm{dw}=2{\int }_1^2\frac{{(\mathrm{w}-1)}^2}{{\mathrm{w}}^5}\mathrm{dw}$$$$\therefore I=2{\int }_1^2\frac{{\mathrm{w}}^2-2\mathrm{w}+1}{{\mathrm{w}}^5}\mathrm{dw}=2{\int }_1^2[{\mathrm{w}}^{-3}-{2\mathrm{w}}^{-4}+{\mathrm{w}}^{-5}]\mathrm{dw}$$$$\Rightarrow I=2{(-\frac{{\mathrm{w}}^{-2}}{2}+\frac{2}{3}{\mathrm{w}}^{-3}-\frac{{\mathrm{w}}^{-4}}{4})}_1^2$$$$\Rightarrow I=2{[-\frac{1}{{2\mathrm{w}}^2}+\frac{2}{{3\mathrm{w}}^3}-\frac{1}{{4\mathrm{w}}^4}]}_1^2=2[(-\frac{1}{2{\left(2\right)}^2}+\frac{2}{3{\left(2\right)}^3}-\frac{1}{4{\left(2\right)}^4})-(\frac{-1}{2{\left(1\right)}^2}+\frac{2}{3{\left(1\right)}^3}-\frac{1}{4{\left(1\right)}^4})]$$$$I=2[\frac{-1}{8}+\frac{2}{24}-\frac{1}{64}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}]=2\left(\frac{5}{192}\right)$$$$\Rightarrow I=\frac{5}{96}\Rightarrow {\int }_0^{\frac{\pi }{4}}\frac{\sqrt{\mathrm{sinx}}}{{(\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}})}^5}\mathrm{dx}=\frac{5}{96}$$$$$$

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