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\int tan^{-1}\alpha .\tan x\; d\alpha \;equals

  • Option 1)

    tan–1\alpha. secx + c

  • Option 2)

    tan–1\alpha.log(secx) + c

  • Option 3)

    tanx {\alpha tan–1\alpha – 0.5 log (1 + \alpha2)} + c

  • Option 4)

    tan\alpha {x tan–1 x – 0.5log (1 + x2)} + c

 

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

 

  

 

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

 

tanx\int ta{n^{ - 1}}\alpha d\alpha = {\rm{ }}tanx\left[ {\alpha {{\tan }^{ - 1}}\alpha - \int {\frac{\alpha }{{1 + {\alpha ^2}}}d\alpha } } \right]

= {\rm{ }}tanx{\rm{ }}[ata{n^{ - 1}}a - \frac{1}{2}\ln \,\left( {1 + {\alpha ^2}} \right)]{\rm{ }} + {\rm{ }}c\;\;

 


Option 1)

tan–1\alpha. secx + c

Option 2)

tan–1\alpha.log(secx) + c

Option 3)

tanx {\alpha tan–1\alpha – 0.5 log (1 + \alpha2)} + c

Option 4)

tan\alpha {x tan–1 x – 0.5log (1 + x2)} + c

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Aadil

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