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\int \cos x\log \left ( \tan \frac{x}{2} \right )dx=

  • Option 1)

    \sin x\log \left ( \tan \frac{x}{2} \right )+c

  • Option 2)

    \sin x\log \left ( \tan \frac{x}{2} \right )-x+c

  • Option 3)

    \sin x\log \left ( \tan \frac{x}{2} \right )+x+c

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learnt

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

 

 

\log \;\left( {\tan \frac{x}{2}} \right) \cdot \sin x - \int {\frac{1}{{\tan \,(x/2)}}} \cdot {\sec ^2}\frac{x}{2} \cdot \frac{1}{2} \cdot \sin x\;dx + c$

= \sin x \cdot \log \;\left( {\tan \frac{x}{2}} \right) - \int {\frac{1}{{2\sin (x/2)\cos (x/2)}}} \cdot \sin x\;dx + c$

= \sin x \cdot \log \;\left( {\tan \frac{x}{2}} \right) - \;\int {dx + c} $= \sin x \cdot \log \;\left( {\tan \frac{x}{2}} \right) - x + c$


Option 1)

\sin x\log \left ( \tan \frac{x}{2} \right )+c

Option 2)

\sin x\log \left ( \tan \frac{x}{2} \right )-x+c

Option 3)

\sin x\log \left ( \tan \frac{x}{2} \right )+x+c

Option 4)

none of these

Posted by

gaurav

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