Question

Evaluate: \(\smallint \frac{{\sin \left( {2{{\tan }^{ - 1}}x} \right)}}{{1 + {x^2}}}\;dx\)
1. \( - \frac{{\cos \left( {4{{\tan }^{ - 1}}x} \right)}}{4} + C\)
2. \( \frac{{\cos \left( {4{{\tan }^{ - 1}}x} \right)}}{4} + C\)
3. \( \frac{{\cos \left( {2{{\tan }^{ - 1}}x} \right)}}{2} + C\)
4. \( - \frac{{\cos \left( {2{{\tan }^{ - 1}}x} \right)}}{2} + C\)
5. None of these

Answer 1

Correct Answer - Option 4 : \( - \frac{{\cos \left( {2{{\tan }^{ - 1}}x} \right)}}{2} + C\)

Concept:

• \(\smallint \sin \left( {ax} \right)\;dx = \; - \frac{{\cos \left( {ax} \right)}}{a} + C\), where c is a constant

Integration by Substitution:

• If the given integration is of the form \(\smallint {\rm{g}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right){\rm{f'}}\left( {\rm{x}} \right){\rm{dx}}\) where \({\rm{g}}\left( {\rm{x}} \right)\) and \({\rm{f}}\left( {\rm{x}} \right)\) are both differentiable functions then we substitute \({\rm{f}}\left( {\rm{x}} \right) = {\rm{u}}\) which implies that \({\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{du}}\).
• Therefore, the integral becomes \(\smallint {\rm{g}}\left( {\rm{u}} \right){\rm{du}}\) which can be solved by general formulas.

Calculation:

Given: \(\smallint \frac{{\sin \left( {2{{\tan }^{ - 1}}x} \right)}}{{1 + {x^2}}}\;dx\)

Let tan-1 x = t

Now by differentiating tan-1 x = t with respect to x we get,

\(\Rightarrow \frac{1}{{1 + {x^2}}}\;dx = dt\)

\(\Rightarrow \smallint \frac{{\sin \left( {2{{\tan }^{ - 1}}x} \right)}}{{1 + {x^2}}}\;dx = \;\smallint \sin 2t\;\)

As we know that, \(\smallint \sin \left( {ax} \right)\;dx = \; - \frac{{\cos \left( {ax} \right)}}{a} + C\)

\(\Rightarrow \smallint \sin 2t\;dt = \; - \frac{{\cos 2t}}{2} + C\)

Now by substituting  tan-1 x = t in the above equation we get

\(\Rightarrow \smallint \frac{{\sin \left( {2{{\tan }^{ - 1}}x} \right)}}{{1 + {x^2}}}\;dx = - \frac{{\cos \left( {2{{\tan }^{ - 1}}x} \right)}}{2} + C\)