MCQ
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}) = $
- A$2{e^{ax}}$
- B$({a^2} + {b^2}){e^{2ax}}$
- ✓${e^{2ax}}$
- D$({a^2} - {b^2}){e^{2ax}}$
✓
Answer
Correct option: C.
${e^{2ax}}$
c
(c)$u = \int_{}^{} {{e^{ax}}\cos bx\,dx} $$ = {e^{ax}}\frac{{\sin bx}}{b} - \frac{a}{b}\int_{}^{} {{e^{ax}}.\sin bx\,dx} $
$ = \frac{{{e^{ax}}\sin bx}}{b} - \frac{a}{b}v$ $ \Rightarrow bu + av = {e^{ax}}\sin bx$..$(i)$
Similarly $bv - au = - {e^{ax}}\cos bx$..$(ii)$
Squaring $(i)$ and $(ii)$ and adding, we get
$({a^2} + {b^2})({u^2} + {v^2}) = {e^{2ax}}$.
(c)$u = \int_{}^{} {{e^{ax}}\cos bx\,dx} $$ = {e^{ax}}\frac{{\sin bx}}{b} - \frac{a}{b}\int_{}^{} {{e^{ax}}.\sin bx\,dx} $
$ = \frac{{{e^{ax}}\sin bx}}{b} - \frac{a}{b}v$ $ \Rightarrow bu + av = {e^{ax}}\sin bx$..$(i)$
Similarly $bv - au = - {e^{ax}}\cos bx$..$(ii)$
Squaring $(i)$ and $(ii)$ and adding, we get
$({a^2} + {b^2})({u^2} + {v^2}) = {e^{2ax}}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $\vec{\text{a}},\vec{\text{b}}$ represent the diagonals of a rhombus, then:
→- $\vec{\text{a}}\times\vec{\text{b}}=\vec{0}$
- $\vec{\text{a}}.\vec{\text{b}}=0$
- $\vec{\text{a}}.\vec{\text{b}}=1$
- $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}$
Let $f$ be function such that $f(x + y) = f(x) + f(y)$ for all $x$ and $y$ and $f(x) = (2x^2 + 3x)g(x)$ for all $x$; where $g(x)$ is continuous and $g(0) = 3.$ Then $f '(x)$ is equal to :-
→The general solution of the differential equation $x d y+y d x=0$ is:
→In figure, which of the following is not true?
→- $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec0$
-
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec0$
-
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec0$
-
$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec0$
A vector parallel to the line of intersection of the plance $\vec{\text{r}}.(3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=1$ and $\vec{\text{r}}.(\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}})=2$ is:
→- $-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
- $2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$
- $-2\hat{\text{i}}-7\hat{\text{j}}+13\hat{\text{k}}$
- $2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
$\int_{}^{} {\left( {\frac{{2 + \sin 2x}}{{1 + \cos 2x}}} \right)\,\,{e^x}dx = } $
→If A and B are two events such that $\text{P(A)}=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then P(B|A) =
→- $\frac{1}{10}$
- $\frac{1}{8}$
- $\frac{7}{8}$
- $\frac{17}{20}$
If $\alpha, \beta, \gamma$ are the direction angles of a vector and $\cos \alpha=\frac{14}{15}, \cos \beta=\frac{1}{3}$, then $\cos \gamma=$
→Let $m$ and $n$ be odd integers such that $0< m < n$. If $f(x) ={x^{\frac{m}{n}}}$ for $x \in R$, then
→$\int_0^\infty {\frac{{x\,dx}}{{(1 + x)(1 + {x^2})}}} = $
→