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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int\large\frac{tan \: x+tan^3x}{1+tan^3x}$

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1 Answer

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Toolbox:
  • $1+\tan^2x=\sec^2x$
  • Form of the partial function $\large\frac{Ax+B}{(x-a)}+\frac{C}{(x-b)}$
Step 1:
$I=\int\large\frac{\tan x+\tan^3xdx}{1+\tan^3x}$
$\quad=\int\large\frac{\tan x(1+\tan^2x)}{1+\tan^3x}$$dx$
$1+\tan^2x=\sec^2x$
$\therefore I=\int\large\frac{\tan x.\sec^2x}{1+\tan^3x}$$dx$
Step 2:
Put $t=\tan x$
$dt=\sec^2xdx$
$I=\int\large\frac{tdt}{1+t^3}$
$(1+t^3)=(1+t)(1-t+t^2)$
$I=\int\large\frac{tdt}{(1+t)(1-t+t^2)}$
Resolving into partial fraction we get,
$\large\frac{t}{(1+t)(1-t+t^2)}=\frac{A}{(1+t)}+\frac{Bx+c}{1-t+t^2}$
$t=A(1-t+t^2)+(Bx+C)(1+t)$
Put $t=-1$
$-1=A(2+1)$
$\Rightarrow A=\large\frac{-1}{3}$
Comparing the coeff of $t^2$ on both sides we get,
$A+B=0$
$\Rightarrow B=\large\frac{1}{3}$
Comparing the constant term we get,
$A+C=0$
$C=\large\frac{1}{3}$
$\therefore \large\frac{t}{1+t^3}=\frac{1}{3(1+t)}+\frac{(1/3)t+1/3}{1-t+t^2}$
Step 3:
$I=\large\frac{-1}{3}\int\large\frac{1}{t+1}$$dt+\large\frac{1}{3}\int\frac{t+1}{1-t+t^2}$$dt$
Multiply and divide $I_2$ by 2
$I=\large\frac{-1}{3}\int\large\frac{1}{t+1}$$dt+\large\frac{1}{6}\int\frac{2t-1+3}{1-t+t^2}$$dt$
$\;\;=-\large\frac{1}{3}$$\log\mid 1+t\mid+\large\frac{1}{6}\int\frac{2t-1}{1-t+t^2}$$dt+\large\frac{3}{6}\int\frac{1}{t^2-t+1}$$dt$
$\;\;=-\large\frac{1}{3}$$\log\mid 1+t\mid+\large\frac{1}{6}$$\log\mid 1-t+t^2\mid+\large\frac{1}{2}.\frac{1}{(t-1/2)^2+(\sqrt 3/2)^2}$$dt$
$\;\;=-\large\frac{1}{3}$$\log\mid 1+t\mid+\large\frac{1}{6}$$\log\mid 1-t+t^2\mid+\large\frac{1}{\sqrt 3}$$\tan^{-1}\big(\large\frac{2t-1}{\sqrt 3}\big)+c$
Step 4:
Substituting for $t$ we get,
$I=-\large\frac{1}{3}$$\log\mid 1+\tan x\mid+\large\frac{1}{6}$$\log\mid 1-\tan x+\tan^2x\mid+\large\frac{1}{\sqrt 3}$$\tan^{-1}\big(\large\frac{2\tan x-1}{\sqrt 3}\big)+c$
answered Oct 2, 2013 by sreemathi.v
 
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