How do you evaluate the integral $\int \frac{sin \sqrt{x}}{\sqrt{x}}$ $int sinsqrtx/sqrtx$ ?

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How do you evaluate the integral $\int \frac{sin \sqrt{x}}{\sqrt{x}}$ $int sinsqrtx/sqrtx$ ?

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Answer 1 · 2017-03-05T15:39:04

$- 2 cos \sqrt{x} + C$ $-2cossqrt(x) + C$

Explanation:

Let $u = \sqrt{x}$ $u = sqrt(x)$ . Then $d u = \frac{1}{2 \sqrt{x}} d x$ $du = 1/(2sqrt(x)) dx$ and $d x = 2 \sqrt{x} d u$ $dx = 2sqrtxdu$ .

$\int \frac{sin u}{u} \cdot 2 u d u$ $int sinu/u * 2udu$

$2 \int sin u d u$ $2int sinu du$

$- 2 cos u + C$ $-2cosu + C$

$- 2 cos \sqrt{x} + C$ $-2cossqrt(x) + C$

Hopefully this helps!

Answer 2 · 2017-06-22T05:57:35

$- 2 cos \sqrt{x} + C$ $-2cossqrtx+C$

Explanation:

The following is an absolutely ridiculous way of arriving at the correct answer. I only did this method because this question was filed under "Trigonometric Substitutions", so I used a trig function in my substitution.

While this worked, using $u = \sqrt{x}$ $u=sqrtx$ is much more straightforward.

Let $x = {sin}^{2} θ$ $x=sin^2theta$ . Then $d x = 2 sin θ cos θ d θ$ $dx=2sinthetacosthetad theta$ and $\sqrt{x} = sin θ$ $sqrtx=sintheta$ .

$I = \int \frac{sin \sqrt{x}}{\sqrt{x}} d x = \int \frac{sin (sin θ)}{sin θ} 2 sin θ cos θ d θ$ $I=intsinsqrtx/sqrtxdx=intsin(sin theta)/sintheta2sinthetacosthetad theta$

$I = 2 \int sin (sin θ) cos θ d θ$ $color(white)I=2intsin(sin theta)costhetad theta$

Letting $t = sin θ$ $t=sintheta$ shows that $d t = cos θ d θ$ $dt=costhetad theta$ :

$I = 2 \int sin t d t = - 2 cos t = - 2 cos (sin θ) = - 2 cos \sqrt{x} + C$ $I=2intsintdt=-2cost=-2cos(sin theta)=-2cossqrtx+C$

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How do you evaluate the integral $\int \frac{sin \sqrt{x}}{\sqrt{x}}$ $int sinsqrtx/sqrtx$ ?

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