By using the substitution \({x^2} = 2\sec \theta \), show that \(\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }} = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c} \).

EITHER

\({x^2} = 2\sec \theta \)

\(2x\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sec \theta \tan \theta \)     M1A1

\(\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }}} \)

\( = \int {\frac{{\sec \theta \tan \theta {\text{d}}\theta }}{{2\sec \theta \sqrt {4{{\sec }^2}\theta  - 4} }}} \)     M1A1

OR

\(x = \sqrt 2 {(\sec \theta )^{\frac{1}{2}}}{\text{ }}\left( { = \sqrt 2 {{(\cos \theta )}^{ - \frac{1}{2}}}} \right)\)

\(\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = \frac{{\sqrt 2 }}{2}{(\sec \theta )^{\frac{1}{2}}}\tan \theta {\text{ }}\left( { = \frac{{\sqrt 2 }}{2}{{(\cos \theta )}^{ - \frac{3}{2}}}\sin \theta } \right)\)     M1A1

\(\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} - 4} }}} \)

\( = \int {\frac{{\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\tan \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\sqrt {4{{\sec }^2}\theta  - 4} }}{\text{ }}\left( { = \int {\frac{{\sqrt 2 {{(\cos \theta )}^{ - \frac{3}{2}}}\sin \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\cos \theta )}^{ - \frac{1}{2}}}\sqrt {4{{\sec }^2}\theta  - 4} }}} } \right)} \)     M1A1

THEN

\( = \frac{1}{2}\int {\frac{{\tan \theta {\text{d}}\theta }}{{2\tan \theta }}} \)     (M1)

\( = \frac{1}{4}\int {{\text{d}}\theta } \)

\( = \frac{\theta }{4} + c\)     A1

\({x^2} = 2\sec \theta  \Rightarrow \cos \theta  = \frac{2}{{{x^2}}}\)     M1

Note:     This M1 may be seen anywhere, including a sketch of an appropriate triangle.

so \(\frac{\theta }{4} + c = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c\)     AG

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