## anonymous 3 years ago Evaluate the intergal

1. anonymous

$\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt$

2. anonymous

My work: $3\int\limits_{}^{}\sin^2(x)\cos^4(t)dt$ $3\int\limits_{}^{}\sin^2(t)\cos^2(t)\cos^2(t)dt$ $3\int\limits_{}^{}\frac{ 1 }{ 2 }(1-\cos(2t))(\frac{ 1 }{ 2 }(1+\cos(2t)))^2dt$ $\frac{ 3 }{ 8 }\int\limits_{}^{}(1-\cos(2t))((1+\cos(2t)))^2dx$ $\frac{ -3 }{ 8 }\int\limits_{}^{}((\cos^3(2t)+\cos^2(2t)+\cos(2t)+1) dt$ $\frac{ -3 }{ 8 }\int\limits_{}^{}((cos^2(2t))(cos(2t))+\cos^2(2t)+\cos(2t)+1))dt$ $\frac{ -3 }{ 8 }\int\limits_{}^{}(\cos^3(2t)+\cos^2(2t)+\cos(2t)+1)dt$

3. anonymous

I don't know what to do after :( .

4. anonymous

The 4th step should be a dt.

5. anonymous

look in the back of your book at the "reduction" formula i think you can make a u sub earlier on though. maybe i am wrong

6. anonymous

Is there no other way?

7. anonymous

I mean I did get a little further.

8. anonymous

$\frac{ -3 }{ 8 }\int\limits\limits_{}^{}(\cos^2(2t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt$

9. anonymous

i would start with $\int\cos^4(x)dx-\int\cos^6(x)dx$ and then look in the back of the book

10. anonymous

$\frac{ -3 }{ 8 }\int\limits\limits\limits_{}^{}((1-\sin^2(t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt$

11. anonymous

Where did that come from?

12. anonymous

there are fomulas that i cannot remember for integrating $$\sin^n(x)$$ and $$\cos^n(x)$$ and i garantee you they are on the back cover of your text

13. anonymous

Yeah, I know. I turn them into the half angles, which I did.

14. anonymous

That stupid cos^2(2t) is getting in the way.

15. anonymous

$$\sin^2(x)=1-\cos^2(x)$$ then multiply out it is easier than reinventing the wheel you are trying to derive the formula, (which is admirable) but it is easier to look it up

16. anonymous

Using another half angle won't help because they are multiplied together which just makes it squared afain,

17. anonymous

again*

18. anonymous

why i find this topic rather dull. almost everything you need is printed on the back cover of your text

19. anonymous

I know. I have to do it though.

20. anonymous

then i really recommend looking up the formula for $$\int\cos^n(x)dx$$

21. anonymous

So I get:$\frac{ -3 }{ 8 }\int\limits_{}^{}(1-\sin^2(2t))\cos(2t)+(1-\sin^2(2t)+\cos(2t)+1)dt$

22. anonymous

Hmm good point.

23. anonymous

Sec.

24. anonymous

Can I ask a dumb question?

25. anonymous

We have that constant out in front. If we break up the intergal into seperate sums do I distribute that constant to all the intergals?

26. anonymous

sure so i can seem dumber if i cannot answer it

27. anonymous

yes (whew)

28. anonymous

ignore the annoying 3, put it in at the end

29. anonymous

Hehe. Thanks. This should be easy now.

30. anonymous

here it is $\int\cos^n(x)dx = \frac{1}{n}\cos(x)\sin(x)+\frac{n-1}{n}\int \cos^{n-2}(x)dx$ use with $$n=4$$ and again with $$n=6$$

31. anonymous

Ohh that's MASSIVELY convenient.

32. anonymous

it is going to be a pain, but much less of a pain than what you were doing

33. anonymous

Why 4 and 6?

34. anonymous

I have 3 and 2.

35. anonymous

$\sin^2(x)\cos^4(x)=(1-\cos^2(x))\cos^4(x)=\cos^4(x)-\cos^6(x)$

36. anonymous

so 4 and 6

37. anonymous

Right.

38. anonymous

Thanks!

39. anonymous

have fun!

40. anonymous

yw