Question

For x² ≠ nπ + 1, n ∈ N (the set of natural numbers), the integral \(\smallint {\rm{x}}\sqrt {\frac{{2{\rm{sin}}\left( {{{\rm{x}}^2} - 1} \right) - {\rm{sin}}2\left( {{{\rm{x}}^2} - 1} \right)}}{{2{\rm{sin}}\left( {{{\rm{x}}^2} - 1} \right) + {\rm{sin}}2\left( {{{\rm{x}}^2} - 1} \right)}}} {\rm{dx}}\) is equal to:

(where c is a constant of integration)
1. \({\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {\frac{1}{2}{\rm{se}}{{\rm{c}}^2}\left( {{{\rm{x}}^2} - 1} \right)} \right| + {\rm{c}}\)
2. \(\frac{1}{2}{\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {{\rm{sec}}\left( {{{\rm{x}}^2} - 1} \right)} \right| + {\rm{c}}\)
3. \(\frac{1}{2}{\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {{\rm{se}}{{\rm{c}}^2}\left( {\frac{{{{\rm{x}}^2} - 1}}{2}} \right)} \right| + {\rm{c}}\)
4. \({\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {{\rm{sec}}\left( {\frac{{{{\rm{x}}^2} - 1}}{2}} \right)} \right| + {\rm{c}}\)

Answer 1

Correct Answer - Option 4 : \({\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {{\rm{sec}}\left( {\frac{{{{\rm{x}}^2} - 1}}{2}} \right)} \right| + {\rm{c}}\)

From question, the integral given can be taken as I:

\(\Rightarrow {\rm{I}} = \smallint {\rm{x}}\sqrt {\frac{{2{\rm{sin}}\left( {{{\rm{x}}^2} - 1} \right) - {\rm{sin}}2\left( {{{\rm{x}}^2} - 1} \right)}}{{2{\rm{sin}}\left( {{{\rm{x}}^2} - 1} \right) + {\rm{sin}}2\left( {{{\rm{x}}^2} - 1} \right)}}} {\rm{dx}}\) 

\(\because \frac{{{\text{x}}^{2}}-1}{2}=\text{ }\!\!\theta\!\!\text{ }\) 

⇒ x² – 1 = 2θ

\(\Rightarrow 2{\rm{x}}\frac{{{\rm{dx}}}}{{{\rm{d\theta }}}} = 2\)

\(\Rightarrow {\rm{x}}\frac{{{\rm{dx}}}}{{{\rm{d\theta }}}} = 1\)

∴ xdx = dθ

\(\Rightarrow {\rm{I}} = \smallint \sqrt {\frac{{2{\rm{sin}}2{\rm{\theta }} - {\rm{sin}}2\left( {2{\rm{\theta }}} \right)}}{{2{\rm{sin}}2{\rm{\theta }} + {\rm{sin}}2\left( {2{\rm{\theta }}} \right)}}} {\rm{d\theta }}\)

\(\Rightarrow {\rm{I}} = \smallint \sqrt {\frac{{2{\rm{sin}}2{\rm{\theta }} - {\rm{sin}}4{\rm{\theta }}}}{{2{\rm{sin}}2{\rm{\theta }} + {\rm{sin}}4{\rm{\theta }}}}} {\rm{d\theta }}\)

∵ [sin 2θ = 2 sin θ cos θ ⇒ sin 4θ = sin 2(2θ) = 2 sin 2θ cos 2θ]

\(\Rightarrow {\rm{I}} = \smallint \sqrt {\frac{{2{\rm{sin}}2{\rm{\theta }} - 2\sin 2{\rm{\theta }}\cos 2{\rm{\theta }}}}{{2{\rm{sin}}2{\rm{\theta }} + 2\sin 2{\rm{\theta }}\cos 2{\rm{\theta }}}}} {\rm{d\theta }}\)

\(\Rightarrow {\rm{I}} = \smallint \sqrt {\frac{{2{\rm{sin}}2{\rm{\theta }}\left( {1 - \cos 2{\rm{\theta }}} \right)}}{{2{\rm{sin}}2{\rm{\theta }}\left( {1 + \cos 2{\rm{\theta }}} \right)}}} {\rm{d\theta }}\)

\(\Rightarrow {\rm{I}} = \smallint \sqrt {\frac{{1 - \cos 2{\rm{\theta }}}}{{1 + \cos 2{\rm{\theta }}}}} {\rm{\;d\theta }}\)

∵ [1 – cos 2θ = 2 sin² θ]

∵ [1 + cos 2θ = 2 cos² θ]

\(\Rightarrow {\rm{I}} = \smallint \sqrt {\frac{{2{{\sin }^2}{\rm{\theta }}}}{{2{{\cos }^2}{\rm{\theta }}}}} {\rm{d\theta }}\)

\(\Rightarrow {\rm{I}} = \smallint \sqrt {{{\tan }^2}{\rm{\theta }}} {\rm{d\theta }}\)

\(\Rightarrow {\rm{I}} = \smallint \left( {\tan {\rm{\theta }}} \right){\rm{d\theta }}\)

⇒ I = log_e |sec θ|+ C

Now, putting the value of θ,

\(\therefore {\rm{I}} = {\log _{\rm{e}}}\left| {\sec \left( {\frac{{{{\rm{x}}^2} - 1}}{2}} \right)} \right| + {\rm{C}}\)