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Integrate \int \frac{dx}{\sin ^{2}x+2\cos^{2}x+2\sin x\cos x }

  • Option 1)

    x + c

  • Option 2)

    \frac{1}{2}\tan^{-1}\left ( 1+x \right )+C

  • Option 3)

    \tan^{-1}\left ( 1+\tan x \right )+C

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learned,

 

Type of Integration by perfect square -

The integral in the form of :

(i)    \int \frac{px+r}{ax^{2}+bx+c}dx     (ii) \int \frac{px+r}{\sqrt{{ax^{2}+bx+c}}}dx

(iii) \int (px+r){\sqrt{{ax^{2}+bx+c}}}dx  

\therefore \int \frac{px+r}{ax^{2}+bx+c}dx=A\frac{\ d.c.\ of\left ( ax^{2}+bx+c \right )}{ax^{2}+bx+c}+B\cdot \frac{1}{ax^{2}+bx+c}  and find integrals by standard formulae

 

- wherein

Working rule.

Let     px+r=A\frac{\mathrm{d} }{\mathrm{d} x}(ax^{2}+bx+c)+B

Find A and B by comparing

 

 

Divide by \cos ^{2}x

\int \frac{\sec ^{2}xdx}{\tan ^{2}x+2+2\tan x}

I=\int \frac{\sec ^{2}xdx}{\left ( \tan x+1 \right )^{2}+1}\: \Rightarrow Put tanx = t, \sec ^{2}xdx=dt

I=\int \frac{dt}{\left ( t+1 \right )^{2}+1}=\tan ^{-1}\left ( \tan x +1 \right )+C


Option 1)

x + c

Option 2)

\frac{1}{2}\tan^{-1}\left ( 1+x \right )+C

Option 3)

\tan^{-1}\left ( 1+\tan x \right )+C

Option 4)

none of these

Posted by

gaurav

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