some intergrated problems

[warsatan]

I got really rusty with intergration. Here are 3 of them

. . .1) \(\displaystyle \L y\, =\, \int{sin^3(x)}\,dx\)
that one i'm not sure how to aproach.

. . .2) \(\displaystyle \L y\, =\, \int{\frac{1}{25\,-\,x^2}}\,dx\)

the result of this one when i used intergration by part i got

. . .\(\displaystyle \L 0.1\ln{(5\,-\,x)}\, +\, 0.1\ln{(5\,+\,x)}\)

and the last one, no idea

. . .3) \(\displaystyle \L y\, =\, \int{\frac{1}{25\,+\,x^2}}\,dx\)

thanks i you can help or give any advice.

 

[galactus]

I got really rusty with intergration. Here are 3 of them

\(\displaystyle y = \L\\\int{sin^3(x)}dx\)
that one i'm not sure how to aproach.

You could use the reduction formula:

\(\displaystyle \L\\\int{sin^{n}(x)}dx=\frac{-1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n}\int{sin^{n-2}(x)}dx\)



\(\displaystyle y = \L\\\int{1/(25-x^2)}dx\)

the result of this one when i used intergration by part i got
.1ln (5-x) + .1ln (5+x)

Try the formula: \(\displaystyle \L\\\int\frac{1}{a^{2}-x^{2}}dx=\frac{-ln(\frac{x-a}{x+a})}{2a}\)

and the last one, no idea

\(\displaystyle y = \L\\\int{1/(25+x^2)}dx\)

Use the formula \(\displaystyle \L\\\int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}tan^{-1}(\frac{x}{a})\)


thanks i you can help or give any advice.

If you feel real brave, try deriving the formulas.