## fortheloveofscience 4 years ago integrate the following

1. fortheloveofscience

$\int\limits_{0}^{\Pi/2} \sin ^{2}x \div (\sin ^{2}x+4\cos ^{2}x)$

2. perl

change cos^2 = 1 - sin^2

3. perl

so you have for the denominator sin^2 + 4(1-sin^2) = sin^2 + 4 - 4 sin^2 -3sin^2 + 4

4. shankvee\

I think this is aprtial fractions. If you take cos^2 x common from denominator you have tan^2 x/(tan^2 x+4) Put tan x=t. you'll get dx=dt/sec^2 x= dt/(1+t^2). So you have t^2/((t^2+4)(t^2+1)). Do you know partial fractions?

5. perl

you dont need partional fractions

6. shankvee\

How will you do it by converting denominator to sin?

7. perl

the denominator is -3sin^2 t + 4 , do long division

8. fortheloveofscience

we gottta use the properties of definite integrals

9. perl

shank, change cos^2 = 1 - sin^2

10. shankvee\

And then what?

11. perl

long division

12. shankvee\

?

13. perl

|dw:1328867815825:dw|

14. shankvee\

You're effectively doing the same thing only.

15. perl

|dw:1328867933423:dw|

16. shankvee\

By doing by partial fractions it reduces to the same thing. Only it is in terms of t^2 so easier to integrate. You directly get tan inverse form which is easy.

17. perl

ok you did trig substitution

18. shankvee\

For the second term you divide by cos ^2 and then you get tan form and then you put that as t to integrate, You get that directly by partial fractions in terms of t.

19. perl

20. perl

what sort of substitution is this called?

21. shankvee\

Yeah thats a trignometric substituion.

22. perl

ok your solution is better, im not sure if i would go ahead with mine . not sure what do do next

23. perl

you can divide top and bottom by cos^2 for my solution

24. perl