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

Evaluate the integral: $\int \operatorname{Tan}^{-1}\left(x^{\frac{1}{3}}\right) d x$

$\int \tan^{-1} x \, dx = $

$\int \frac{\sin^{-1}x}{(1-x^2)^{3/2}} \, dx = $

If $\int(\log x)^3 x^5 d x=\frac{x^6}{A}\left[B(\log x)^3+C(\log x)^2+D(\log x)-1\right]+k$ and $A, B, C, D$ are integers, then $A-(B+C+D)=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module