integration question?

Question

1.integrate [(3 /sqrt (1 + 9y^2)) dy]

2.integrate (sqrt ( y^2 - 25)) / y^3 dy, y > 5

I missed these two problems on the test and I want to get it right, So I can learn it for the finals. But I have no time to meet with my professor, so I need your help. I'd appreciate it very much

Thank you.

Answer 1

Ok, the firs integration is :

1) integrate( 3 / sqrt( 1 + 9y^2))dy

this is like :

3*integrate(dy / 9y^2 + 1)

integrate(dy / 9y^2 + 1) >>> That's a known integration and has this form :

1/3*arctan(3y)

then : 3*integrate(dy / 9y^2 + 1)) :

arctan(3y)

2) integrate(sqrt( y^2 - 25 / y^3)dy)

Here you need to make a substitution :

y^2 - 25 / y^2 = t^2

1 - 25y^-2 = t^2 >>>> 50y^-3dy = 2tdt >>>> tdt / 25 = y^-3dy

y^2 - 25 = y^2*t^2

replacing it on the integration :

integrate((tdt / 25)*(y^2-25)^1/2)

integrate((tdt /25)*5t / sqrt(1 - t^2))

1/5*integrate(dt*t^2 / sqrt(1 - t^2))

let's call : t = sin(a) >>> dt = cos(a)d(a)

1/5*integrate(cos(a)*sin^2(a) /cos(a))

1/5*integrate(sin^2(a)d(a)

well that's more simple :

1/5*integrate( 1 - cos(2a) / 2) d(a)

1/10*integrate[d(a) - cos(2a)da]

1/10*[ a - 1/2sin(a)]

remembering : t = sin(a) and 1 - 25y^-2 = t^2

1/10*[arcsin(t) - 1/2t]

1/10*[arcsin(sqrt(1 - 25y^-2) - 1/2*sqrt(1-25y^-2)]

That's it.

The last one was very long, but hope you can understand it.

Answer 2

It is better that u apply yr own brain. Help in mathematics is not advisable. it is a subject of practice once u have known the basics

Answer 3

Integration by skill of areas. The function is a made of two purposes: (x) * (cosx) call considered one of them "u" and the different "dv" choose the "u" so as that its spinoff simplifies u = x du = a million*dx dv = cosx v = sinx it particularly is for the formula setup, it is u*v - vital (vdu) =xsinx - vital(sinx * a million * dx) =xsinx - (-cosx) + C = xsinx + cosx + C