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If \int \frac{\tan x}{1+\tan x +\tan^{2}x}dx=

x- \frac{K}{\sqrt{A}}\tan ^{-1} (\frac{K\tan x+1} {\sqrt{A}})+c

(C is a constant of integration), then the ordered
pair (K, A) is equal to :

 

  • Option 1)

    (2, 1)

  • Option 2)

    (−2, 3)

  • Option 3)

    (2, 3)

  • Option 4)

    (−2, 1)

 

Answers (2)

best_answer

As we have learned

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

 

 

- wherein

Since \int f(x)dx=\int f(t)dt=\int f(\theta )d\theta all variables must be converted into single variable ,\left ( t\, or\ \theta \right )

 

 

 

Integration by perfect square method -

ax^{2}+bx+c=a\left [ x^{2}+\frac{bx}{a}+\frac{c}{a} \right ]=a\left [ \left ( x+\frac{b}{2a} \right )^{2}+\frac{c}{a}-\frac{b^{2}}{4a^{2}} \right ]

- wherein

Make the coefficient of x^{2} +ve one . 

 

  

put   \tan x = t ; \sec^{2}x = dt

\int \frac{t}{(1+t+t^{2})(1+t^{2})}dt= \int \frac{1}{1+t^{2}}- \frac{1}{(1+t+t^{2})}dt

\tan ^{-1}(t)- \int \frac{dt}{(t+1/2)^{2}+ (\sqrt3/2)^{2}}

\tan ^{-1}(t)- \frac{1}{\sqrt3/2} \tan \frac{2t+1}{\sqrt3}

thus k= 2  and A= 3

 

 


Option 1)

(2, 1)

This is incorrect

Option 2)

(−2, 3)

This is incorrect

Option 3)

(2, 3)

This is correct

Option 4)

(−2, 1)

This is incorrect

Posted by

Himanshu

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