## fortheloveofscience Group Title integrate the following 2 years ago 2 years ago

1. fortheloveofscience Group Title

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

2. perl Group Title

change cos^2 = 1 - sin^2

3. perl Group Title

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

4. shankvee\ Group Title

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 Group Title

you dont need partional fractions

6. shankvee\ Group Title

How will you do it by converting denominator to sin?

7. perl Group Title

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

8. fortheloveofscience Group Title

we gottta use the properties of definite integrals

9. perl Group Title

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

10. shankvee\ Group Title

And then what?

11. perl Group Title

long division

12. shankvee\ Group Title

?

13. perl Group Title

|dw:1328867815825:dw|

14. shankvee\ Group Title

You're effectively doing the same thing only.

15. perl Group Title

|dw:1328867933423:dw|

16. shankvee\ Group Title

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 Group Title

ok you did trig substitution

18. shankvee\ Group Title

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 Group Title

20. perl Group Title

what sort of substitution is this called?

21. shankvee\ Group Title

Yeah thats a trignometric substituion.

22. perl Group Title

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

23. perl Group Title

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

24. perl Group Title