If u = int e^{ax} cos bx , dx and v = int e^{ | vedclass

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

Question

(C) Using integration by parts for $u = \int e^{ax} \cos bx \, dx$: $u = e^{ax} \frac{\sin bx}{b} - \int a e^{ax} \frac{\sin bx}{b} \, dx = \frac{e^{a

Vedclass · Accepted Answer

$e^{2ax}$

Vedclass · Answer

(C) Using integration by parts for $u = \int e^{ax} \cos bx \, dx$: $u = e^{ax} \frac{\sin bx}{b} - \int a e^{ax} \frac{\sin bx}{b} \, dx = \frac{e^{ax} \sin bx}{b} - \frac{a}{b}v$ Multiplying by $b$,we get $bu + av = e^{ax} \sin bx$ ... $(i)$ Similarly,for $v = \int e^{ax} \sin bx \, dx$: $v = e^{ax} \left( -\frac{\cos bx}{b} \right) - \int a e^{ax} \left( -\frac{\cos bx}{b} \right) \, dx = -\frac{e^{ax} \cos bx}{b} + \frac{a}{b}u$ Multiplying by $b$,we get $bv - au = -e^{ax} \cos bx$ ... $(ii)$ Squaring and adding equations $(i)$ and $(ii)$: $(bu + av)^2 + (bv - au)^2 = (e^{ax} \sin bx)^2 + (-e^{ax} \cos bx)^2$ $b^2u^2 + a^2v^2 + 2abuv + b^2v^2 + a^2u^2 - 2abuv = e^{2ax} (\sin^2 bx + \cos^2 bx)$ $(a^2 + b^2)u^2 + (a^2 + b^2)v^2 = e^{2ax}(1)$ $(a^2 + b^2)(u^2 + v^2) = e^{2ax}$

If $u = \int e^{ax} \cos bx \, dx$ and $v = \int e^{ax} \sin bx \, dx$,then $(a^2 + b^2)(u^2 + v^2) = $

Similar Questions

If $\int f(x) dx = \varphi(x)$,then $\int x^5 f(x^3) dx = $

Integrate the function: $\left(x^{2}+1\right) \log x$

$\int {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)\,dx$

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module