x Turn on thread page Beta
 You are Here: Home >< Maths

# Intergration of Trig watch

1. Can someone please show how to do these type of questions?

Intergrate cos^4(x)sin^3(x) dx
2. (Original post by Dopey')
Can someone please show how to do these type of questions?

Intergrate cos^4(x)sin^3(x) dx
Well with that one you use sin^2x = 1 - cos^2x

Then you have 2 parts that look like cos^nxsinx so you use inverse chain rule
3. (Original post by TenOfThem)
Well with that one you use sin^2x = 1 - cos^2x

Then you have 2 parts that look like cos^nxsinx so you use inverse chain rule
cos^4(x)sinx(sin^2(x))
cos^4(x)sinx(1-cos(x))]

By the way its not cos to the power of 4x its cos to the power of 4
4. (Original post by Dopey')
cos^4(x)sinx(sin^2(x))
cos^4(x)sinx(1-cos(x))]

By the way its not cos to the power of 4x its cos to the power of 4
Yeah I know

Your working above is incorrect sin^2x=1-cos^x

Then multiply the bracket
5. (Original post by TenOfThem)
Yeah I know

Your working above is incorrect sin^2x=1-cos^x

Then multiply the bracket
cos^4(x)sinx(1-cos^2x)
cos^4(x)sinx -cos^6(x)sinx

Its this ok so far, what the next step?

think ive got it

i did

cos^4(x)sinx(1-cos^2(x)) then u = cos x u' = -sinx dx/du = -cosec x

put u in

- u^4 + u^6 then it grate and substitute cos x back in after
6. (Original post by Dopey')
cos^4(x)sinx(1-cos^2x)
cos^4(x)sinx -cos^6(x)sinx

Its this ok so far, what the next step?

think ive got it

i did

cos^4(x)sinx(1-cos^2(x)) then u = cos x u' = -sinx dx/du = -cosec x

put u in

- u^4 + u^6 then it grate and substitute cos x back in after
Will the same technique apply to intergrating sin^3x dx ?

my steps so far

Sinx(sin^2x)
Sinx(1-cos^2x) Now I cant see what to do?
7. (Original post by Dopey')
Will the same technique apply to intergrating sin^3x dx ?

my steps so far

Sinx(sin^2x)
Sinx(1-cos^2x) Now I cant see what to do?
Expand the bracket. Now integrate term by term (you can use a substitution for the second term).
8. (Original post by Mr M)
Expand the bracket. Now integrate term by term (you can use a substitution for the second term).
I think im getting the hang of it, but it always seems to get harder could you show how you would solve this please

Intergrate in(rt(x-1))dx
9. (Original post by Dopey')
I think im getting the hang of it, but it always seems to get harder could you show how you would solve this please

Intergrate in(rt(x-1))dx
Why does everyone think the natural logarithm is written as "in"?

Make a substitution

Remember than

Integrate by parts.
10. (Original post by Mr M)
Why does everyone think the natural logarithm is written as "in"?

Make a substitution

Remember than

Integrate by parts.
Haha I dunno why everyone does, It just one of those things i guess

One more question I'm attempting to intergrate x^5e^(x^3)

I intergrate by parts but it just goes round in circle (x keeps coming back) is there a special technique for these type of questions?
11. (Original post by Dopey')
Haha I dunno why everyone does, It just one of those things i guess

One more question I'm attempting to intergrate x^5e^(x^3)

I intergrate by parts but it just goes round in circle (x keeps coming back) is there a special technique for these type of questions?
You just need lots of experience.

Write as and make a substitution .
12. (Original post by Mr M)
You just need lots of experience.

Write as and make a substitution .
so is dv/dx = x^2e^(x^3)?

v = 1/3e^(x^3)?

I think Ive done it the answers right but the workings looks horrible, maybe theres a quicker method?

u = x^3

u' = 3x^2

v = 1/3u v' = 1/3
dz/dx = e^u z= e^u

= 1/3ue^u - (intergrate) 1/3e^u
= 1/3ue^u - 1/3e^u
= 1/3x^3e(x^3) - 1/3e^(x^3) = 1/3e^(x^3)(x^3-1) + c

Is this ok?
13. (Original post by Dopey')
so is dv/dx = x^2e^(x^3)?

v = 1/3e^(x^3)?
You seem to have failed to make the substitution I suggested?
14. (Original post by Dopey')
so is dv/dx = x^2e^(x^3)?

v = 1/3e^(x^3)?
That'll work.
15. (Original post by Mr M)
You seem to have failed to make the substitution I suggested?
Ive a adjusted my workings above ^ (how you suggested)
16. (Original post by Dopey')
the workings looks horrible, maybe theres a quicker method?
Maybe MAKING A SUBSTITUTION would help?

17. (Original post by Mr M)
Maybe MAKING A SUBSTITUTION would help?

I did substitute u = x^3 in, If you didnt mean it like this i dont understand im a novice
18. (Original post by Dopey')
I did substitute u = x^3 in, If you didnt mean it like this i dont understand im a novice
My fault I think. The choice of u was confusing for you as you muddled it with u from integration by parts. This is why I changed it to t above.
19. (Original post by Mr M)
Maybe MAKING A SUBSTITUTION would help?

To be fair to Dopey, he saw your breakdown of x^5 with u=x^3 and thought you were doing IBP with dv/dx equal the rest.
20. (Original post by notnek)
To be fair to Dopey, he saw your breakdown of x^5 with u=x^3 and thought you were doing IBP with dv/dx equal the rest.
Yes I realise that. I failed to communicate properly.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 14, 2013
Today on TSR

### Boyfriend slept with someone else

...we were on a break

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams