Question #7d3ed

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Question #7d3ed

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Answer 1 · 2017-12-06T18:33:34

$\sqrt{2} arcsin (\sqrt{2} sin x) - arctan (\frac{sin x}{\sqrt{cos 2 x}}) + C$ $sqrt2arcsin(sqrt2sinx)-arctan(sinx/sqrt(cos2x))+C$

Explanation:

$\int \frac{\sqrt{cos 2 x}}{cos x} \cdot d x$ $int sqrt(cos2x)/cosx*dx$

= $\int \frac{\sqrt{1 - 2 {(sin x)}^{2}}}{{(cos x)}^{2}} \cdot cos x \cdot d x$ $int sqrt[1-2(sinx)^2]/(cosx)^2*cosx*dx$

= $\int \frac{\sqrt{1 - 2 {(sin x)}^{2}}}{1 - {(sin x)}^{2}} \cdot cos x \cdot d x$ $int sqrt[1-2(sinx)^2]/[1-(sinx)^2]*cosx*dx$

After using $\sqrt{2} sin x = sin y$ $sqrt2sinx=siny$ an $\sqrt{2} cos x \cdot d x = cos y \cdot d y$ $sqrt2cosx*dx=cosy*dy$ substitution, this integral became

= $\int \frac{\sqrt{1 - {(sin y)}^{2}}}{1 - {(\frac{sin y}{\sqrt{2}})}^{2}} \cdot (\frac{cos y \cdot d y}{\sqrt{2}})$ $int sqrt[1-(siny)^2]/[1-(siny/sqrt2)^2]*((cosy*dy)/sqrt2)$

= $\sqrt{2} \int \frac{\sqrt{{(cos y)}^{2}}}{2 - {(sin y)}^{2}} \cdot cos y \cdot d y$ $sqrt2int sqrt[(cosy)^2]/[2-(siny)^2]*cosy*dy$

= $\sqrt{2} \int \frac{{(cos y)}^{2} \cdot d y}{2 - {(sin y)}^{2}}$ $sqrt2int ((cosy)^2*dy)/[2-(siny)^2]$

= $\sqrt{2} \int \frac{d y}{2 {(sec y)}^{2} - {(tan y)}^{2}}$ $sqrt2int (dy)/[2(secy)^2-(tany)^2]$

= $\sqrt{2} \int \frac{d y}{2 {(tan y)}^{2} + 2 - {(tan y)}^{2}}$ $sqrt2int (dy)/[2(tany)^2+2-(tany)^2]$

= $\sqrt{2} \int \frac{d y}{{(tan y)}^{2} + 2}$ $sqrt2int (dy)/[(tany)^2+2]$

= $\sqrt{2} \int \frac{[{(tan y)}^{2} + 1] \cdot d y}{[{(tan y)}^{2} + 1] [{(tan y)}^{2} + 2]}$ $sqrt2int ([(tany)^2+1]*dy)/([(tany)^2+1][(tany)^2+2])$

After using $z = tan y$ $z=tany$ and $d z = [{(tan y)}^{2} + 1] \cdot d y$ $dz=[(tany)^2+1]*dy$ transformation, it became

$\sqrt{2} \int \frac{d z}{(z^{2} + 1) \cdot (z^{2} + 2)}$ $sqrt2int (dz)/[(z^2+1)*(z^2+2)]$

= $\sqrt{2} \int \frac{d z}{z^{2} + 1}$ $sqrt2int (dz)/(z^2+1)$ - $\sqrt{2} \int \frac{d z}{z^{2} + 2}$ $sqrt2int (dz)/(z^2+2)$

= $\sqrt{2} arctan z - arctan (\frac{z}{\sqrt{2}}) + C$ $sqrt2arctanz-arctan(z/sqrt2)+C$

= $\sqrt{2} arctan (tan y) - arctan (\frac{tan y}{\sqrt{2}}) + C$ $sqrt2arctan(tany)-arctan(tany/sqrt2)+C$

= $y \sqrt{2} - arctan (\frac{tan y}{\sqrt{2}}) + C$ $ysqrt2-arctan(tany/sqrt2)+C$

After using $\sqrt{2} sin x = sin y$ $sqrt2sinx=siny$ , $y = arcsin (\sqrt{2} sin x)$ $y=arcsin(sqrt2sinx)$ and $tan y = \frac{\sqrt{2} sin x}{\sqrt{cos 2 x}}$ $tany=(sqrt2sinx)/sqrt(cos2x)$ inverse transforms, I found

$\int \frac{\sqrt{cos 2 x}}{cos x} \cdot d x$ $int sqrt(cos2x)/cosx*dx$

= $\sqrt{2} arcsin (\sqrt{2} sin x) - arctan (\frac{sin x}{\sqrt{cos 2 x}}) + C$ $sqrt2arcsin(sqrt2sinx)-arctan(sinx/sqrt(cos2x))+C$

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Question #7d3ed

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