Evaluate int sin (sqrt{k}) , dk on (0, infty) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int \sin (\sqrt{k}) \, dk$ for $k \in (0, \infty)$.Substitute $k = t^2$,which implies $dk = 2t \, dt$.Substituting these into the integr

Vedclass · Accepted Answer

$2[\sin (\sqrt{k})-\sqrt{k} \cos (\sqrt{k})]+c$

Vedclass · Answer

(D) Let $I = \int \sin (\sqrt{k}) \, dk$ for $k \in (0, \infty)$.Substitute $k = t^2$,which implies $dk = 2t \, dt$.Substituting these into the integral,we get:$I = \int \sin(t) \cdot (2t \, dt) = 2 \int t \sin(t) \, dt$.Using integration by parts,where $\int u \, dv = uv - \int v \, du$,let $u = t$ and $dv = \sin(t) \, dt$. Then $du = dt$ and $v = -\cos(t)$.$I = 2 [t(-\cos(t)) - \int (-\cos(t)) \, dt]$$I = 2 [-t \cos(t) + \int \cos(t) \, dt]$$I = 2 [-t \cos(t) + \sin(t)] + c$Substituting $t = \sqrt{k}$ back into the expression:$I = 2[\sin(\sqrt{k}) - \sqrt{k} \cos(\sqrt{k})] + c$.

Evaluate $\int \sin (\sqrt{k}) \, dk$ on $(0, \infty)$.

Similar Questions

The integral $\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$ is equal to (where $c$ is a constant of integration)

$\int \frac{dx}{(a^2 + x^2)^{3/2}}$ is equal to

$\int {\frac{t}{e^{3t^2}}} \, dt = $

$\int \frac{(5 \sin \theta-2) \cos \theta}{(5-\cos ^2 \theta-4 \sin \theta)} d \theta=$

Let $f(x) = \int x^3 \sqrt{3-x^2} dx$. If $5f(\sqrt{2}) = -4$,then $f(1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module