Integration

masterjoint · #1 $Old$ November 24th, 2008, 02:35 AM

there are 3 some questions from my Calculus homework that are giving me real head ache ! , please help dudes

have to get the integral of the followings
1. dx / ( 2e^x ) * sqr root ( 1 - e^2x ) INDEFINITE

2. { sqr root [ (1 - x^2)^3 ] } dx [0,1] DEFINITE

3. { x^3 * sin^2 * x / x^4 + 2x^2 + 1 } dx [-5,5] DEFINITE

please write the way of how to do each question with formulas used, it would be very very nice of you ! thanks alot in advanced !

ADARSH · #2 $Old$ November 24th, 2008, 05:53 AM

Hello & a very warm Welcome to the forum

Quote:

Originally Posted by masterjoint $View Post$

1. dx / ( 2e^x ) * sqr root ( 1 - e^2x )

$I=\int{\frac{dx}{2e^x * \sqrt{1-e^{2x}}}}$

put $e^x=t$

so $dx=\frac{dt}{t}$

hence
$I=\int{\frac{dt}{2t^2 * (1-t^2)^{1/2}}}$

so

$I=\int{\frac{dt}{2t^3 * \sqrt{(\frac{1}{t^2} - 1)}}}$

put $\frac{1}{t^2}= k$

so
$-2t^{-3} dt = dk$

hence
$I=\int { \frac{dk}{-4*\sqrt{k-1}}}$

put $(k-1)=g$

so
$I=\int { \frac{dg}{-4*\sqrt{g}}}$

I am too lazy to continue

so go ahead if unsolved feel free to ask

galactus · #3 $Old$ November 24th, 2008, 06:12 AM

The second one is difficult to integrate by elementary methods. But, I do not think it is required. It is a matter of observation to note we have an odd function. Since we integrate from -5 to 5, it evaluates to 0.