int cos sqrt{x} , dx = (where C is a constant | vedclass

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

Question

(A) To evaluate the integral $I = \int \cos \sqrt{x} \, dx$,we use the substitution method. Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To evaluate the integral $I = \int \cos \sqrt{x} \, dx$,we use the substitution method. Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$. Substituting these into the integral,we get: $I = \int \cos(t) \cdot (2t \, dt) = 2 \int t \cos t \, dt$. Now,we apply integration by parts,where $\int u \, dv = uv - \int v \, du$. Let $u = t$ and $dv = \cos t \, dt$. Then $du = dt$ and $v = \sin t$. $I = 2 [t \sin t - \int \sin t \, dt]$. $I = 2 [t \sin t - (-\cos t)] + C$. $I = 2 [t \sin t + \cos t] + C$. Substituting $t = \sqrt{x}$ back into the expression: $I = 2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C$.

$\int \cos \sqrt{x} \, dx =$ (where $C$ is a constant of integration.)

Similar Questions

$\int \frac{e^{\tan ^{-1} 2 x}}{1+4 x^2} dx =$

$\int \frac{\sin x}{3+4 \cos ^2 x} d x$

$\int \frac{x^{\frac{1}{3}}}{(1 + x^{\frac{2}{3}})^3} dx$ is equal to (where $C$ is the constant of integration).

$\int \frac{dx}{1 + 3\sin^2 x} = $

If $\int \frac{dx}{5 + 4\cos x} = \lambda \tan^{-1} \left( m \tan \frac{x}{2} \right) + C$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module