## Dido525 2 years ago Evaluate the intergal

1. Dido525

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

2. Dido525

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

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

4. Dido525

The 4th step should be a dt.

5. satellite73

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

Is there no other way?

7. Dido525

I mean I did get a little further.

8. Dido525

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

9. satellite73

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

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

11. Dido525

Where did that come from?

12. satellite73

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

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

14. Dido525

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

15. satellite73

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

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

17. Dido525

again*

18. satellite73

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

19. Dido525

I know. I have to do it though.

20. satellite73

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

21. Dido525

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

22. Dido525

Hmm good point.

23. Dido525

Sec.

24. Dido525

Can I ask a dumb question?

25. Dido525

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

sure so i can seem dumber if i cannot answer it

27. satellite73

yes (whew)

28. satellite73

ignore the annoying 3, put it in at the end

29. Dido525

Hehe. Thanks. This should be easy now.

30. satellite73

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

Ohh that's MASSIVELY convenient.

32. satellite73

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

33. Dido525

Why 4 and 6?

34. Dido525

I have 3 and 2.

35. satellite73

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

36. satellite73

so 4 and 6

37. Dido525

Right.

38. Dido525

Thanks!

39. satellite73

have fun!

40. satellite73

yw