Question Number 107069 by mathdave last updated on 08/Aug/20
Answered by bemath last updated on 09/Aug/20
$$\:\:\:\:\:\:\:@{bemath}@ \\ $$$$\left(\sqrt{\mathrm{cos}\:{x}}\:+\sqrt{\mathrm{sin}\:{x}}\right)^{\mathrm{5}} =\left(\sqrt{\mathrm{cos}\:{x}}\left(\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\right)^{\mathrm{5}} \right. \\ $$$$=\:\mathrm{cos}\:^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}\:{x}}\:\left(\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\right)^{\mathrm{5}} \\ $$$${I}=\:\int\:\frac{\sqrt{\mathrm{tan}\:{x}}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}}{\left(\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\right)^{\mathrm{5}} }\: \\ $$$${set}\:\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\:=\:{z}\:\Rightarrow\sqrt{\mathrm{tan}\:{x}}\:=\:{z}−\mathrm{1} \\ $$$$\frac{\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}}{\mathrm{2}\sqrt{\mathrm{tan}\:{x}}}\:=\:{dz}\:\Rightarrow{dx}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{tan}\:{x}}\:{dz}}{\mathrm{sec}\:^{\mathrm{2}} {x}} \\ $$$${I}\:=\:\int\:\frac{\sqrt{\mathrm{tan}\:{x}}\:\mathrm{sec}\:^{\mathrm{2}} {x}}{{z}^{\mathrm{5}} }.\left(\frac{\mathrm{2}\sqrt{\mathrm{tan}\:{x}}}{\mathrm{sec}\:^{\mathrm{2}} {x}}\:{dz}\right) \\ $$$${I}=\:\int\:\frac{\mathrm{2}\left({z}−\mathrm{1}\right)^{\mathrm{2}} {dz}}{{z}^{\mathrm{5}} } \\ $$$${I}=\mathrm{2}\int\:\frac{{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{5}} }\:{dz}\:=\:\mathrm{2}\int{z}^{−\mathrm{3}} −\mathrm{2}{z}^{−\mathrm{4}} +{z}^{−\mathrm{5}} \:{dz} \\ $$$${I}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}{z}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}{z}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}{z}^{\mathrm{4}} }\right)+{C} \\ $$$${I}=\:−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{3}{z}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}{z}^{\mathrm{4}} }+{C} \\ $$$${I}=\:\frac{−\mathrm{6}{z}^{\mathrm{2}} +\mathrm{8}{z}−\mathrm{3}}{\mathrm{6}{z}^{\mathrm{4}} }+{C}\: \\ $$$${I}=\:\frac{−\mathrm{6}\left(\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\right)^{\mathrm{2}} +\mathrm{8}\sqrt{\mathrm{tan}\:{x}}+\mathrm{5}}{\mathrm{6}\left(\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\right)^{\mathrm{4}} }+{C} \\ $$
Commented by bobhans last updated on 09/Aug/20
![I = [((−6(1+(√(tan x)))^2 +8(√(tan x))+5)/(6(1+(√(tan x)))^4 )) ]_0 ^(π/4) I= [((−6(4)+8+5)/(6.16))]−[((−6+5)/6)] I= −((11)/(96))+(1/6)= ((−11+16)/(96)) = (5/(96))](Q107079.png)
$$\mathrm{I}\:=\:\left[\frac{−\mathrm{6}\left(\mathrm{1}+\sqrt{\mathrm{tan}\:\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{8}\sqrt{\mathrm{tan}\:\mathrm{x}}+\mathrm{5}}{\mathrm{6}\left(\mathrm{1}+\sqrt{\mathrm{tan}\:\mathrm{x}}\right)^{\mathrm{4}} }\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\mathrm{I}=\:\left[\frac{−\mathrm{6}\left(\mathrm{4}\right)+\mathrm{8}+\mathrm{5}}{\mathrm{6}.\mathrm{16}}\right]−\left[\frac{−\mathrm{6}+\mathrm{5}}{\mathrm{6}}\right] \\ $$$$\mathrm{I}=\:−\frac{\mathrm{11}}{\mathrm{96}}+\frac{\mathrm{1}}{\mathrm{6}}=\:\frac{−\mathrm{11}+\mathrm{16}}{\mathrm{96}}\:=\:\frac{\mathrm{5}}{\mathrm{96}} \\ $$
Answered by Tony6400 last updated on 08/Aug/20
![Evaluate ∫_0 ^(π/4) (((√(sinx))dx)/(((√(cosx))+(√(sinx)))^5 )) Take ((√(cosx))+(√(sinx)))^5 =[(√(cosx))(1+((√(sinx))/(√(cosx))))]^5 =[(√(cosx))]^4 .(√(cosx))[1+(√(tanx))]^5 =cos^2 x(√(cosx))(1+(√(tanx)))^5 ∴I=∫_0 ^(π/4) ((√(sinx))/(cos^2 x(√(cosx))(1+(√(tanx)))^5 ))dx ⇒I=∫_0 ^(π/4) (((√(tanx)).sec^2 x)/((1+(√(tanx)))^5 ))dx Let w=1+(√(tanx))⇒(dw/dx)=((sec^2 x)/(2(√(tanx))))⇒dx=((2(√(tanx)))/(sec^2 x))dw When x=(π/4),w=2.When x=0,w=1⇒I=∫_1 ^2 (((√(tanx)).sec^2 x)/w^5 ).((2(√(tanx)))/(sec^2 x))dw I=2∫_1 ^2 ((tanx)/w^5 )dw=2∫_1 ^2 (((w−1)^2 )/w^5 )dw ∴I=2∫_1 ^2 ((w^2 −2w+1)/w^5 )dw=2∫_1 ^2 [w^(−3) −2w^(−4) +w^(−5) ]dw ⇒I=2(−(w^(−2) /2)+(2/3)w^(−3) −(w^(−4) /4))_1 ^2 ⇒I=2[−(1/(2w^2 ))+(2/(3w^3 ))−(1/(4w^4 ))]_1 ^2 =2[(−(1/(2(2)^2 ))+(2/(3(2)^3 ))−(1/(4(2)^4 )))−(((−1)/(2(1)^2 ))+(2/(3(1)^3 ))−(1/(4(1)^4 )))] I=2[((−1)/8)+(2/(24))−(1/(64))+(1/2)−(2/3)+(1/4)]=2((5/(192))) ⇒I=(5/(96))⇒∫_0 ^(π/4) ((√(sinx))/(((√(cosx))+(√(sinx)))^5 ))dx=(5/(96))](Q107111.png)
$$\mathrm{Evaluate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{sinx}}\mathrm{dx}}{\left(\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}}\right)^{\mathrm{5}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{Take}\:\left(\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}}\right)^{\mathrm{5}} =\left[\sqrt{\mathrm{cosx}}\left(\mathrm{1}+\frac{\sqrt{\mathrm{sinx}}}{\sqrt{\mathrm{cosx}}}\right)\right]^{\mathrm{5}} =\left[\sqrt{\mathrm{cosx}}\right]^{\mathrm{4}} .\sqrt{\mathrm{cosx}}\left[\mathrm{1}+\sqrt{\mathrm{tanx}}\right]^{\mathrm{5}} \\ $$$$=\mathrm{cos}^{\mathrm{2}} \mathrm{x}\sqrt{\mathrm{cosx}}\left(\mathrm{1}+\sqrt{\mathrm{tanx}}\right)^{\mathrm{5}} \\ $$$$\therefore{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{sinx}}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}\sqrt{\mathrm{cosx}}\left(\mathrm{1}+\sqrt{\mathrm{tanx}}\right)^{\mathrm{5}} }\mathrm{dx} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{tanx}}.\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{1}+\sqrt{\mathrm{tanx}}\right)^{\mathrm{5}} }\mathrm{dx} \\ $$$$\mathrm{Let}\:\mathrm{w}=\mathrm{1}+\sqrt{\mathrm{tanx}}\Rightarrow\frac{\mathrm{dw}}{\mathrm{dx}}=\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{2}\sqrt{\mathrm{tanx}}}\Rightarrow\mathrm{dx}=\frac{\mathrm{2}\sqrt{\mathrm{tanx}}}{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}\mathrm{dw} \\ $$$$\mathrm{When}\:\mathrm{x}=\frac{\pi}{\mathrm{4}},\mathrm{w}=\mathrm{2}.\mathrm{When} \\ $$$$\mathrm{x}=\mathrm{0},\mathrm{w}=\mathrm{1}\Rightarrow{I}=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\sqrt{\mathrm{tanx}}.\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{w}^{\mathrm{5}} }.\frac{\mathrm{2}\sqrt{\mathrm{tanx}}}{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}\mathrm{dw} \\ $$$${I}=\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{tanx}}{\mathrm{w}^{\mathrm{5}} }\mathrm{dw}=\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left(\mathrm{w}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{w}^{\mathrm{5}} }\mathrm{dw} \\ $$$$\therefore{I}=\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{w}^{\mathrm{2}} −\mathrm{2w}+\mathrm{1}}{\mathrm{w}^{\mathrm{5}} }\mathrm{dw}=\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{w}^{−\mathrm{3}} −\mathrm{2w}^{−\mathrm{4}} +\mathrm{w}^{−\mathrm{5}} \right]\mathrm{dw} \\ $$$$\Rightarrow{I}=\mathrm{2}\left(−\frac{\mathrm{w}^{−\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{w}^{−\mathrm{3}} −\frac{\mathrm{w}^{−\mathrm{4}} }{\mathrm{4}}\right)_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow{I}=\mathrm{2}\left[−\frac{\mathrm{1}}{\mathrm{2w}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3w}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4w}^{\mathrm{4}} }\right]_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}\left[\left(−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}\right)^{\mathrm{4}} }\right)−\left(\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}\right)^{\mathrm{4}} }\right)\right] \\ $$$${I}=\mathrm{2}\left[\frac{−\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{2}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{64}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right]=\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{192}}\right) \\ $$$$\Rightarrow{I}=\frac{\mathrm{5}}{\mathrm{96}}\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{sinx}}}{\left(\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}}\right)^{\mathrm{5}} }\mathrm{dx}=\frac{\mathrm{5}}{\mathrm{96}} \\ $$$$ \\ $$