int cos sqrt{x} , dx = | vedclass

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

Question

(A) Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.Substituting these into the integral:$\int \cos \sqrt{x} \, dx = \int \cos(t) \cd

Vedclass · Accepted Answer

$2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + c$

Vedclass · Answer

(A) Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.Substituting these into the integral:$\int \cos \sqrt{x} \, dx = \int \cos(t) \cdot 2t \, dt = 2 \int t \cos(t) \, dt$.Using integration by parts,where $u = t$ and $dv = \cos(t) \, dt$:$\int u \, dv = uv - \int v \, du = t \sin(t) - \int \sin(t) \, dt = t \sin(t) + \cos(t)$.Multiplying by the constant $2$:$2(t \sin(t) + \cos(t)) + c$.Substituting $t = \sqrt{x}$ back:$2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + c$.

$\int \cos \sqrt{x} \, dx = $

Similar Questions

If $\int {\ln ({x^2} + x)dx = x\ln ({x^2} + x) + A}$,then $A = $

Let $\int x^3 \sin x \, dx = g(x) + C$,where $C$ is the constant of integration. If $8\left(g\left(\frac{\pi}{2}\right) + g^{\prime}\left(\frac{\pi}{2}\right)\right) = \alpha \pi^3 + \beta \pi^2 + \gamma$,where $\alpha, \beta, \gamma \in \mathbb{Z}$,then $\alpha + \beta - \gamma$ equals:

$\int x \log x \, dx$ is equal to

If $\int x(1+x) \log(1+x^2) dx = F(x) \log(1+x^2) - \frac{2}{3} \tan^{-1} x - \frac{2x^3}{9} - \frac{x^2}{2} + \frac{2x}{3} + c$, then $F(x) =$

The value of the integral $\int x^3 \cos x \, dx$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module