## Dido525 Group Title Evaluate the intergal one year ago one year ago

1. Dido525 Group Title

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

2. Dido525 Group Title

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. Dido525 Group Title

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

4. Dido525 Group Title

The 4th step should be a dt.

5. satellite73 Group Title

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. Dido525 Group Title

Is there no other way?

7. Dido525 Group Title

I mean I did get a little further.

8. Dido525 Group Title

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

9. satellite73 Group Title

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

10. Dido525 Group Title

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

11. Dido525 Group Title

Where did that come from?

12. satellite73 Group Title

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. Dido525 Group Title

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

14. Dido525 Group Title

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

15. satellite73 Group Title

$$\sin^2(x)=1-\cos^2(x)$$ 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. Dido525 Group Title

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

17. Dido525 Group Title

again*

18. satellite73 Group Title

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

19. Dido525 Group Title

I know. I have to do it though.

20. satellite73 Group Title

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

21. Dido525 Group Title

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

22. Dido525 Group Title

Hmm good point.

23. Dido525 Group Title

Sec.

24. Dido525 Group Title

Can I ask a dumb question?

25. Dido525 Group Title

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. satellite73 Group Title

sure so i can seem dumber if i cannot answer it

27. satellite73 Group Title

yes (whew)

28. satellite73 Group Title

ignore the annoying 3, put it in at the end

29. Dido525 Group Title

Hehe. Thanks. This should be easy now.

30. satellite73 Group Title

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. Dido525 Group Title

Ohh that's MASSIVELY convenient.

32. satellite73 Group Title

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

33. Dido525 Group Title

Why 4 and 6?

34. Dido525 Group Title

I have 3 and 2.

35. satellite73 Group Title

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

36. satellite73 Group Title

so 4 and 6

37. Dido525 Group Title

Right.

38. Dido525 Group Title

Thanks!

39. satellite73 Group Title

have fun!

40. satellite73 Group Title

yw