Math Help - Help integration problem

1. Help integration problem

the question is (y^2)/(25+y^2) dy.

here are my steps:
a=5
y=5tanx
dy=5secx^2 dx

[(5tanx)^2 * 5secx^2]/[25*(1+tanx^2)] dx
= [(5tanx)^2 * 5secx^2]/[25*secx^2] dx
=5tanx^2 dx
im stuck here.

I tried to use integration by parts to solve 5tanx^2 dx. but it didn't work out

The final answer should be y-5arctan(y/5)+C which c is any constant

2. Originally Posted by questionboy
the question is (y^2)/(25+y^2) dy.

here are my steps:
a=5
y=5tanx
dy=5secx^2 dx

[(5tanx)^2 * 5secx^2]/[25*(1+tanx^2)] dx
= [(5tanx)^2 * 5secx^2]/[25*secx^2] dx
=5tanx^2 dx
im stuck here.

I tried to use integration by parts to solve 5tanx^2 dx. but it didn't work out

The final answer should be y-5arctan(y/5)+C which c is any constant

y^2/(y^2+25) = [1 - 25/(y^2+25)] = [ 1 - 1/(1 +(y/5)^2)], and you have the sum of two almost immediate integrals

Tonio

3. Originally Posted by questionboy
the question is (y^2)/(25+y^2) dy.

here are my steps:
a=5
y=5tanx
dy=5secx^2 dx

[(5tanx)^2 * 5secx^2]/[25*(1+tanx^2)] dx
= [(5tanx)^2 * 5secx^2]/[25*secx^2] dx
=5tanx^2 dx
im stuck here.

I tried to use integration by parts to solve 5tanx^2 dx. but it didn't work out

The final answer should be y-5arctan(y/5)+C which c is any constant
Try setting $5 \tan^2 x = 5 (\sec^2 x - 1)$ Integrating $\sec^2$ is easy because of what the differential of $\tan$ is ...

4. Originally Posted by Matt Westwood
Try setting $5 \tan^2 x = 5 (\sec^2 x - 1)$ Integrating $\sec^2$ is easy because of what the differential of $\tan$ is ...
thank you very much

5(secx^2-1) dx =5tanx-5x
which x= arctan(y/5)
5tanx-5arctan(y/5) since y=5tanx
y-5arctan(y/5)

I got it...Thank you very much