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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\int\large\frac{2+\sin2x}{1+\cos2x}$$e^x$

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Toolbox:
  • $\int e^x[f(x)+f'(x)]dx=e^x[f(x)]+c$
  • $\frac{1}{\cos^2x}=\sec^2x$
  • $\sin 2x=2\sin x\cos x$
  • $ 1+\cos 2x=2 \cos ^2x$
Step 1:
Given $\large\int\frac{2+\sin2x}{1+\cos2x}$$e^x$
Consider $2+\sin 2x$
we know $\sin 2x=2\sin x \cos x$
Therefore $2+2\sin x\cos x=2(1+\sin x\cos x)$
Consider $1+ \cos 2x$
we know $\cos 2x=2 \cos^2x-1$
Therefore $ 1+\cos 2x=1+2 \cos ^2x-1=2 \cos ^2x$
Therefore $ I=\int\large \frac{2(1+\sin x \cos x)}{2(\cos^2x)}dx$
Step 2:
On seperating the terms we get ,
$I=\int e^x\bigg[\large\frac{1}{\cos ^2x}+\frac{\sin x \cos x}{\cos ^2x}\bigg]$$dx$
But we know $\large\frac{1}{\cos ^2x}=$$\sec ^2x$ and $\large\frac{\sin x}{\cos x}=$$\tan x$
Hence $I=\int e^x[\tan x+\sec^2]dx$
Step 3:
Let $f(x)=\tan x$
Therefore $f'(x)=\sec ^2x$
we know $\int e^x[f(x)+f'(x)]dx=e^x(f(x))+c$
Therefore $ I=e^x \tan x+c$
answered Sep 12, 2013 by sreemathi.v
 
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