Integral

dhiab

Calculate :
$image=http://latex.codecogs.com/gif.latex?\120dpi+\large+\int+\left+(+\frac{x}{xsinx+cosx}+\right+)^{2}dx&hash=298dfaa537fb25a3bb1c0d0cdcbfff7c$

TheCoffeeMachine

simplependulum

MHF Hall of Honor
Calculate :
$image=http://latex.codecogs.com/gif.latex?\120dpi+\large+\int+\left+(+\frac{x}{xsinx+cosx}+\right+)^{2}dx&hash=298dfaa537fb25a3bb1c0d0cdcbfff7c$

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
Method 2 :

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

The integral

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

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

$$\displaystyle = -\frac{x}{ \cos{x}} \left( \frac{1}{x\sin{x} + \cos{x}}\right) + \tan{x} + C$$

AllanCuz

Note that $$\displaystyle x \sin{x} + \cos{x} = \sqrt{x^2+1} \cos( x- \tan^{-1}(x) )$$
Where did this come from?

Thanks

Last edited:

TheCoffeeMachine

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 $$\displaystyle \alpha$$.

AllanCuz

Random Variable

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

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

$$\displaystyle = x \sin x + \cos x$$

dhiab and AllanCuz

AllanCuz

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)

TheCoffeeMachine