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

The value of $\int \sin \sqrt{x} \, dx$ is equal to

If $\int {{x^5}{e^{ - 4{x^3}}}\,dx = \frac{1}{{48}}{e^{ - 4{x^3}}}f\left( x \right) + C} $,where $C$ is a constant of integration,then $f(x)$ is equal to

$\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x = ?$ (where $|x| < 1$)

Find $\int \log x \, dx$.

$\int \frac{\cot^{-1}(e^x)}{e^x} dx$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module