Evaluate: int u frac{d^2v}{dx^2} dx - int v f | vedclass

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

Question

(A) Using integration by parts for both integrals: $\int u \frac{d^2v}{dx^2} dx = u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx$ $\int v \frac

Vedclass · Accepted Answer

$u \frac{dv}{dx} - v \frac{du}{dx} + C$

Vedclass · Answer

(A) Using integration by parts for both integrals: $\int u \frac{d^2v}{dx^2} dx = u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx$ $\int v \frac{d^2u}{dx^2} dx = v \frac{du}{dx} - \int \frac{dv}{dx} \frac{du}{dx} dx$ Subtracting the two expressions: $\int u \frac{d^2v}{dx^2} dx - \int v \frac{d^2u}{dx^2} dx = (u \frac{dv}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx) - (v \frac{du}{dx} - \int \frac{dv}{dx} \frac{du}{dx} dx) + C$ $= u \frac{dv}{dx} - v \frac{du}{dx} - \int \frac{du}{dx} \frac{dv}{dx} dx + \int \frac{du}{dx} \frac{dv}{dx} dx + C$ $= u \frac{dv}{dx} - v \frac{du}{dx} + C$

Evaluate: $\int u \frac{d^2v}{dx^2} dx - \int v \frac{d^2u}{dx^2} dx$

Similar Questions

$\int \frac{x + \sin x}{1 + \cos x} \, dx$ is equal to

When $x > 0$,then $\int \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) dx$ is

If $\int e^{x^2} \cdot x^3 \, dx = e^{x^2} f(x) + c$ and $f(1) = 0$ (where $c$ is a constant of integration),then the value of $f(x)$ is

If $f(x) = \frac{\sin^{-1} x}{\sqrt{1-x^2}}$ and $g(x) = e^{\sin^{-1} x}$, then the value of $\int f(x)g(x) dx = \dots$

If $\int \log \left(a^2+x^2\right) d x=h(x)+C$,then $h(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module