# Evaluate: $\smallint \frac{{dx}}{{x \;\times\; {{\cos }^2}\left( {1 \;+\; \log x} \right)}}$

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

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Correct Answer - Option 3 : $\tan \left( {1 + \log x} \right) + C$

Concept:

• $\smallint {\sec ^2}x\;dx = \tan x + c$, 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: $\smallint \frac{{dx}}{{x\; \times\; {{\cos }^2}\left( {1\; +\; \log x} \right)}}$

Let 1 + log x = t

Now by differentiating the above equation with respect to t we get,

$\Rightarrow \frac{{d\left( {1 + \log x} \right)}}{{dx}} = \frac{{dt}}{{dx}}$

$\Rightarrow \frac{1}{x} \cdot dx = dt$

$\Rightarrow \smallint \frac{{dx}}{{x{{\cos }^2}\left( {1 + \log x} \right)}} = \;\smallint \frac{{dt}}{{{{\cos }^2}t}} = \;\smallint {\sec ^2}t\;dt$

As we know that, $\smallint {\sec ^2}x\;dx = \tan x + c$ , where c is a constant

$\Rightarrow \smallint {\sec ^2}t\;dt = \tan t + C$

Now by substituting 1 + log x = t in the above equation we get,

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