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

1. 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.

Thank you...
2. Factorise. Make a substitution. What do you see?
3. (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.

Thank you...
Can you post all your working?
4. (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.

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

You should know an identity involving tan squared that you can use.
5. (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.
Attached Images

6. (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.
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.
7. (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 ?
8. (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.
Mistake on 4th line.

Note that:

9. Not
10. (Original post by ztibor)
Do you think ?
11. (Original post by Bushido Brown)
Mistake on 4th line.

Note that:

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...
12. To everyone here who tried to help me :THANK YOU

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

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