C4 Integration

So I've had a sheet of 50 integration questions to do and several on the sheet have so far stumped me.

Any help would be appreciated thanks!

9) Sinxcos^(4)x

13) Sin^(2)2x

18) x/9x^2 + 1

24) cot^(2)3x

31) (x+1)^2/x^2+1.
I used division to work out it was the integral of 1 +(2x/x^2+1) Then how do i progress further?

Reply 1

username878045

17

Original post by Henry.Lister

So I've had a sheet of 50 integration questions to do and several on the sheet have so far stumped me.

Any help would be appreciated thanks!

9) Sinxcos^(4)x

13) Sin^(2)2x

18) x/9x^2 + 1

24) cot^(2)3x

31) (x+1)^2/x^2+1.
I used division to work out it was the integral of 1 +(2x/x^2+1) Then how do i progress further?

For many of them, you have to consider the chain rule in reverse (when you have a function multiplied by it's differential). Others involve natural logs. Does this help?

Reply 2

Henry.Lister

OP

Hmm i recognise where you're coming from but I'm weary about knowing exactly how to do it.

Mind showing me one? Thanks

Reply 3

username878045

17

Original post by Henry.Lister

Hmm i recognise where you're coming from but I'm weary about knowing exactly how to do it.

Mind showing me one? Thanks

For 9), consider the differential of cos^(5)x.

For 18), consider the differential of ln(9x^2 +1)

Reply 4

ztibor

10

Original post by Henry.Lister

So I've had a sheet of 50 integration questions to do and several on the sheet have so far stumped me.

Any help would be appreciated thanks!

9) Sinxcos^(4)x

13) Sin^(2)2x

18) x/9x^2 + 1

24) cot^(2)3x

31) (x+1)^2/x^2+1.
I used division to work out it was the integral of 1 +(2x/x^2+1) Then how do i progress further?

For 9)
Use u=cosx substitution
du/dx=-sinx -> -sinx dx =du
or use directly the rule of

\int f'(x)\cdot f^n(x) dx =\frac{f^{n+1}(x)}{n+1}+C

that is

\int sinx \cdot cos^4xdx=-\int (-\sin x)\cdot (\cos x)^4 dx

for 13)
Use the identity of

\sin^2 A=\frac{1-\cos 2A}{2}

and
when the primitive function for f(x) is F(x)+C
then the primitive function for f(ax+b) is F(ax+b)/a +C

for 18)

\int \frac{f'(x)}{f(x)} dx =\ln |f(x)| +C

For this quesion

\int \frac{x}{9x^2+1} dx=\frac{1}{18}\int \frac{18x}{9x^2+1} dx

But you can use substitution of t=9x^2+1
then dt=18x dx -> x dx =1/18 dt

\int \frac{x}{9x^2+1} dx=\int \frac{1}{9x^2+1} \cdot x \cdot dx=\int \frac{1}{t}\cdot \frac{1}{18} dt

for 24)

cot^2 3x=\frac{\cos^2 3x}{\sin^2 3x}=\frac{1-sin^2 3x}{sin^2 3x}=\frac{1}{sin^2 3x}-1

And as you know

-\int \frac{1}{\sin^2 (ax+b)} dx =\frac{\cot (ax+b)}{a} +C

for 31)
see 18

(edited 11 years ago)

Reply 5

username878045

17

Original post by ztibor

x

You're not supposed to give solutions on these forums, but to help the OP reach the answer themselves :smile:

Reply 6

Henry.Lister

OP

9 makes sense now thanks. I'll look through 18 next.

Reply 7

Henry.Lister

OP

Ztibor thanks, I'll look through those calculations shortly :smile:

Reply 8

Henry.Lister

OP

I think I've managed to complete them all except I'm unsure about 13.

I tried changing the integral of sin^22x into (1-cos4x)/2 using cos2x=1-2sin^2x

I then integrated that to get 1/2x-1/4sin4x+c.

Reply 9

joostan

13

Original post by Henry.Lister

I think I've managed to complete them all except I'm unsure about 13.

I tried changing the integral of sin^22x into (1-cos4x)/2 using cos2x=1-2sin^2x

I then integrated that to get 1/2x-1/4sin4x+c.

This is the correct method - but you've made a small slip up in the solution.

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C4 Integration

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