# Thread: Integration

1. ## Integration

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 !

2. ## ...

Hello & a very warm Welcome to the forum
Originally Posted by masterjoint
1. dx / ( 2e^x ) * sqr root ( 1 - e^2x )
$\displaystyle I=\int{\frac{dx}{2e^x * \sqrt{1-e^{2x}}}}$

put $\displaystyle e^x=t$

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

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

so

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

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

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

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

put $\displaystyle (k-1)=g$

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

I am too lazy to continue so go ahead if unsolved feel free to ask

3. 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.