The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

How to integrate (tan x)^(n+2) + (tan x) ^n

HI,

this really tricky question popped up during my practice exam and believe me is the first one of its kind that I would have to seek your kind help for:

Please consider the question in the picture and explain in detail (in step by step manner). How to solve it.

math_image.jpg

Thank you...
Reply 1
Avatar for BabyMaths
BabyMaths
Factorise. Make a substitution. What do you see?
Reply 2
Avatar for Notnek
Notnek
21
Original post by methewthomson
HI,

this really tricky question popped up during my practice exam and believe me is the first one of its kind that I would have to seek your kind help for:

Please consider the question in the picture and explain in detail (in step by step manner). How to solve it.

math_image.jpg

Thank you...

Can you post all your working?
Original post by methewthomson
HI,

this really tricky question popped up during my practice exam and believe me is the first one of its kind that I would have to seek your kind help for:

Please consider the question in the picture and explain in detail (in step by step manner). How to solve it.

math_image.jpg

Thank you...


It's not really tricky at all. Do what you usually do with the substitution, and note that

tan⁡n+2x=tan⁡nx×tan⁡2x\tan^{n+2} x = \tan^n x \times \tan^2 x

You should know an identity involving tan squared that you can use.
(edited 11 years ago)
Original post by notnek
Can you post all your working?

Thanks everyone for replying.

Here is my working in this picture, so please magnify it by clicking on it. As you can see, my answer turns out to be zero while it is not the correct answer. Please consider my working and point out where I am wrong....thanks.:smile:
image.jpg
image.jpg118.9KB
Reply 5
Avatar for CocoPop
CocoPop
18
Original post by methewthomson
Thanks everyone for replying.

Here is my working in this picture, so please magnify it by clicking on it. As you can see, my answer turns out to be zero while it is not the correct answer. Please consider my working and point out where I am wrong....thanks.:smile:
image.jpg


Haven't read the whole thing, but for starters sec^2(x) = 1 + tan^2(x), not 1 - tan^2(x)...

EDIT: More pointers.

If you factorise tan^(n+2)(x) + tan^n(x) = tan^n(x)[1 + tan^2(x)], and you know that 1 + tan^2(x) = sec^2(x) = du/dx, then:

(tan^(n+2)(x) + tan^n(x))dx = u*du. Now you've got an easy integral.
(edited 11 years ago)
Reply 6
Avatar for ztibor
ztibor
10
Original post by CocoPop
Haven't read the whole thing, but for starters sec^2(x) = 1 + tan^2(x), not 1 - tan^2(x)...

EDIT: More pointers.

If you factorise tan^(n+2)(x) + tan^n(x) = tan^n(x)[1 + tan^2(x)], and you know that 1 + tan^2(x) = sec^2(x) = du/dx, then:

(tan^(n+2)(x) + tan^n(x))dx = u*du. Now you've got an easy integral.


Do you think un⋅duu^n\cdot du ?
Original post by methewthomson
Thanks everyone for replying.

Here is my working in this picture, so please magnify it by clicking on it. As you can see, my answer turns out to be zero while it is not the correct answer. Please consider my working and point out where I am wrong....thanks.:smile:
image.jpg


Mistake on 4th line.


Note that:

∫un+1du=u(n+1)+1n+1+1+K=un+2n+2+K\displaystyle \int u^{n+1} du = \dfrac{u^{(n+1)+1}}{n+1+1} + K = \dfrac{ u^{n+2}}{n+2} + K
Reply 8
Avatar for j4ckd4v13z
j4ckd4v13z
sin2x+cos2x=1 sin^2x+cos^2x = 1

⇒tan2x+1=sec2x\Rightarrow tan^2x + 1 = sec^2x

Not 1−tan2x 1-tan^2x
Reply 9
Avatar for CocoPop
CocoPop
18
Original post by ztibor
Do you think un⋅duu^n\cdot du ?


Yup, my bad.
Original post by Bushido Brown
Mistake on 4th line.


Note that:

∫un+1du=u(n+1)+1n+1+1+K=un+2n+2+K\displaystyle \int u^{n+1} du = \dfrac{u^{(n+1)+1}}{n+1+1} + K = \dfrac{ u^{n+2}}{n+2} + K



Thank you so very much for actually pointing out the big error....you have helped me a lot. I suppose now I will solve the whole question correctly...
To everyone here who tried to help me :THANK YOU:smile:

It was indeed very kind of you all....

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you