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  • Help me solve this - The integralis equal to : (where C is a constant of integration) - Integral Calculus - JEE Main

The integral \int \cos \left ( \log_{e}x \right )dx is equal to : 

(where C is a constant of integration) 

  • Option 1)

    x\left [ \cos \left ( \log_{e}x \right ) +\sin \left ( \log_{e}x \right )\right ]+C

     

     

     

  • Option 2)

    \frac{x}{2}\left [ \cos \left ( \log_{e}x \right ) +\sin \left ( \log_{e}x \right )\right ]+C

  • Option 3)

    \frac{x}{2}\left [ \sin \left ( \log_{e}x \right ) -\cos \left ( \log_{e}x \right )\right ]+C

  • Option 4)

    x\left [ \cos \left ( \log_{e}x \right ) -\sin \left ( \log_{e}x \right )\right ]+C

Answers (1)

best_answer

 

Integration By PARTS -

Let u and v are two functions then 

\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx

- wherein

Where u is the Ist function v is he IInd function

 

 

 

 

Rule for integration by parts -

Take Ist function (u) as according I L A T E

  

- wherein

Where ,

I : Inverse

L : Logarithmic

A : Algebraic 

T : Trignometric

E : Exponential

Using by parts 

I=x\cos (\log x)+\int \frac{x}{x}\sin (\log x)dx

   =x\cos (\log x)+\frac{1}{2}[x\sin (\log x)-\int \cos (\log x)dx]+C

  =\frac{x}{2}(\cos (\log x)+\sin (\log x))+C

 

  


Option 1)

x\left [ \cos \left ( \log_{e}x \right ) +\sin \left ( \log_{e}x \right )\right ]+C

 

 

 

Option 2)

\frac{x}{2}\left [ \cos \left ( \log_{e}x \right ) +\sin \left ( \log_{e}x \right )\right ]+C

Option 3)

\frac{x}{2}\left [ \sin \left ( \log_{e}x \right ) -\cos \left ( \log_{e}x \right )\right ]+C

Option 4)

x\left [ \cos \left ( \log_{e}x \right ) -\sin \left ( \log_{e}x \right )\right ]+C

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