Integral

simplependulum

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Jan 2009
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Calculate :

Note that \(\displaystyle x \sin{x} + \cos{x} = \sqrt{x^2+1} \cos( x- \tan^{-1}(x) )\)

We have

\(\displaystyle = \int \frac{x^2}{ (x^2+1) \cos^2( x - \tan^{-1}(x) )}~dx \)

Sub. \(\displaystyle x - \tan^{-1}(x) = t \)

\(\displaystyle (1 - \frac{1}{x^2+1} )dx = dt \)

\(\displaystyle \frac{x^2}{x^2+1}~dx = dt \)

The integral becomes

\(\displaystyle \int \sec^2(t)~dt = \tan(t) + C \)

\(\displaystyle = \tan[ x - \tan^{-1}(x) ] + C \) or

\(\displaystyle = \frac{ \tan(x) - x }{ 1 + x\tan(x) } + C \)
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
Method 2 :

We have \(\displaystyle \frac{d}{dx}( x\sin{x} + \cos{x} ) = x\cos{x} \)

The integral

\(\displaystyle = \int \frac{x}{ \cos{x}} \frac{ x\cos{x}}{ (x\sin{x} + \cos{x})^2}~dx \)

\(\displaystyle = -\frac{x}{ \cos{x}} \left( \frac{1}{x\sin{x} + \cos{x}} \right)+ \int \frac{1}{\cos^2{x}}~dx \)

\(\displaystyle = -\frac{x}{ \cos{x}} \left( \frac{1}{x\sin{x} + \cos{x}}\right) + \tan{x} + C \)
 
Mar 2010
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381
Where did this come from?
By writing \(\displaystyle x\sin{x}+\cos{x}\) in the form \(\displaystyle R\cos\left(x-\alpha\right)\). Let \(\displaystyle x\sin{x}+\cos{x} = R\cos\left(x-\alpha\right)\), expand the right-hand-side, and compare the coefficients and then use Pythagoras' theorem to find \(\displaystyle R\) and the right-angle relationship to find \(\displaystyle \alpha\).
 
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May 2009
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\(\displaystyle \sqrt{ x^{2}+1} \cos(x - \arctan x) = \sqrt{x^{2}+1} \big( \cos x \cos (\arctan x) + \sin x \sin (\arctan x) \big) \)

\(\displaystyle = \sqrt{x^{2}+1} \big( \cos x \frac{1}{\sqrt{x^{2}+1}}+ \sin x \frac{x}{\sqrt{x^{2}+1}} \big) \)

\(\displaystyle = x \sin x + \cos x \)
 
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Apr 2010
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Canada
By writing \(\displaystyle x\sin{x}+\cos{x}\) in the form \(\displaystyle R\cos\left(x-\alpha\right)\). Let \(\displaystyle x\sin{x}+\cos{x} = R\cos\left(x-\alpha\right)\), expand the right-hand-side, and compare the coefficients and then use pythagoras' theorem to find \(\displaystyle R\) and the right-angle relationship to find \(\displaystyle \alpha\).
\(\displaystyle Rcos(x-a) = R[ cosxcosa +sinxsina ] = [R cos(a)]cos(x) + [Rsin(a)]sin(x) \)

\(\displaystyle Rcos(a) = 1 \) and \(\displaystyle Rsin(a) = x\)

\(\displaystyle tan(a) = x \) \(\displaystyle \to a = tan^{-1} (x) \)

The triangle is then,

\(\displaystyle Hyp = \sqrt{ x^2 + 1^2 } \)

Where

\(\displaystyle R = \frac{x}{sin(a)} = \frac{x}{ \frac{x}{ \sqrt{ x^2 + 1^2 } } } = \sqrt{ x^2 + 1^2 } \)

Yup! Thanks, perhaps i need more coffee this morning (Happy)
 
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