Question

integral of (sin^3xcos^7x) dx

Answer 1

This problem might be easier for you as well as for potential helpers if you'd type or write it in as \[\sin^3x \cos^7x\]


Hint: Think of a common identity for (cos x)^2.

Answer 2

Keep one cos x factor and convert the rest of (cos x)^6 into terms involving sin x only.

Answer 3

Im guessing I can use the half angle formula for \[\cos^2x = (1+cos2x)/2 ?\]

Answer 4

Will that convert (cos x)^6 into sine and constant terms only?

Answer 5

or is it just \[1+cos2x?\]

Answer 6

Think: Pythagorean Identity applied to (cos x)^2.

Answer 7

\[(1-sin^2)^5?\]

Answer 8

excuse me, that's \[\sin^2x + \cos^2x=1\]

Answer 9

yes, sorry I was remembering the identity

Answer 10

You're getting there. But why the exponent 5?

Answer 11

Express\[\cos^6 x\] in terms of sin x and constants only.

Answer 12

I thought taking out the \[sin^2x\] from \[\sin^7x\] would leave the exponent 5?

Answer 13

\[\sin^3x * \cos^7x=\sin^3 x \cos^6x \cos x\]

Answer 14

Now think of a replacement for (cos x)^2 in terms of 1 and sin x only:

Answer 15

\[\cos^2x=1-\sin^2x\]

Answer 16

Note that \[\cos^6x=(\cos^2x)^3\]

Answer 17

Oh, so you factored out a single cosx first to put it in terms of sin?

Answer 18

Leave the initial (sin x)^3 as is. Replace (cos x)^2 with 1- (sin x)^2, INSIDE parentheses.

Answer 19

Cube the quantity within parentheses:

Answer 20

\[(\cos x)^2 cubed=(1-\sin^2 x)^3\]

Answer 21

Ohh ok. I think that is what messed up my answer. I kept trying to factor out a cos^2x at the beginning so that I could use the half angle formula

Answer 22

Remember...you still have a factor of (sin x)^3 in front of that.

Answer 23

We learn by doing. Happy to be of help.

Can you finish this integral? I have time to stay with you if you like.

Answer 24

\[(\sin^3 x)*(1-\sin^2 x)^3 * \cos x = ?\]

Answer 25

Hint:\[(a+b)^3=a^3+3a^2b+3ab^2+b^3\]

Answer 26

and\[(a-b)^3=a^3-3a^2b+3ab^2-b^3\]

Answer 27

I am simplifying now...

Answer 28

In your shoes I'd say: "I am now expanding\[(1-\sin^2x)^3.\]

Answer 29

\[\sin^3x(1^3-3*1^2*\sin^2 x + 3*1^1[\sin^2x]^2 + ??\]

Answer 30

Put a right parenthesis at the end of that. )

Answer 31

\[(-70cos2x-20cos4x+5cos6x+5cos8x+cos10x)/5120\]

Answer 32

^ Plus the constant. Sorry I meant I was simplifying my final step.

Answer 33

What I'd do would start out like this:\[\sin^3x*1 -\sin^3x*3\sin^2x + .... \]

Answer 34

so I'm wondering where the coefficients 70, -20, 5, 5, ... came from.

Answer 35

Please expand the \[\sin^3x(1^3-3*1^2*\sin^2 x + 3*1^1[\sin^2x]^2 + ??\] before integrating.

Answer 36

My mistake, integrated the cosin incorrectly.

Answer 37

Complicated. To integrate (sin x)^3, you'd have to use that same trig identity agaion.

Answer 38

Rodr: forgive me, but I need to get off the 'Net for 30 min. I'll be back and would gladly continue with you if you still need me then.

Answer 39

\[\cos^(10)x/10-\cos^8 x/8 +C \]

Answer 40

Note that some calculus books have tables of integrals, and you might be lucky enough to find a formula for the integral of (sin x)^3.

I'll be back! But I'm leaving my computer for the moment.

Answer 41

It's ok, I think I got the correct answer. Thanks for the help and patience:] . And my apologies, my internet is very slow at times :/

Answer 42

Just so that you know: Some calculus books have detailed tables of integrals. I've just found one that applies. It's for\[\sin^n ax *\cos^m ax *dx\]

Answer 43

...but it's important to know how to integrate as you and I have done.

Answer 44

From Mathematica 9:
\[\frac{1}{512} (-7) \cos (2 x)-\frac{1}{256} \cos (4 x)+\frac{\cos (6 x)}{1024}+\frac{\cos (8 x)}{1024}+\frac{\cos (10 x)}{5120} \]