logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Evaluate the following problems using second fundamental theorem: $\int\limits_{0}^{\large\frac{\pi}{2}} e^{3x}\cos x dx$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If $F(x)=\int \limits_a^x f(t)dt $ then $\int \limits_a^b f(x) dx=F(b)-F(a)$
$\int \limits_0^{\large\frac{\pi}{2}} e^{3x} \cos x dx$
Step 1:
Let $ I= \int e^{3x} \cos x=\int udv$
Where $u= \cos x\;and\; dv=e^{3x}dx$
$du=-\sin x \;dx\; and \;v=\large\frac{e^{3x}}{3}$
$I= uv-\int vdu$
$\quad= \cos x \large\frac{e^{3x}}{3}+\int \large\frac {e^3x}{3}$$ \sin x dx$
Step 2:
Let $ I_1= \large\frac{1}{3} \int e^{3x}$$ \sin x dx =\int udv$
Where $u= \sin x,\; dv=e^{3x}dx$
$du=\cos x \;dx,\;v=\large\frac{e^{3x}}{3}$
$I_1= uv-\int vdu$
$\quad= \large\frac{e^{3x}}{9}$$\sin x -\large\frac{1}{9} \int$$ e^3x \cos x dx$
$\quad= \large\frac{e^{3x}}{9}$$\sin x -\large\frac{1}{9}$$ I$
Step 3:
$\therefore I=\cos x \large\frac{e^{3x}}{3}+\frac{e^{3x}}{9}$$ \sin x-\large\frac{1}{9}$$ I$
$\large\frac{10}{9}$$I=\large\frac{e^{3x}}{3}$$ [\cos x+\large\frac{\sin x}{3}]$
$I=\large\frac{3}{10}$$e^{3x} [\cos x+\large\frac{\sin x}{3}]$
$\int \limits_0^{\large\frac{\pi}{2}} e^{3x} \cos x dx=\bigg[\large\frac{3}{10}$$ e^3x [\cos x +\large\frac{\sin x}{3}]\bigg]_0^{\large\frac{\pi}{2}}$
$\qquad=\large\frac{1}{10} e^{\large\frac{3 \pi}{2}}-\large\frac{3}{10}$
$\qquad=\large\frac{1}{10} \bigg(e^{\large\frac{3 \pi}{2}}$$-3\bigg)$
answered Aug 14, 2013 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...