Integration question with power of n

[thoradicus]

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_w11_qp_33.pdf

Homework Statement

no. 10i
cant do ii and iii without doing i first

Homework Equations

tan(x)
d/dx(tan(x))=sec^2x
1+tan^2x=sec^2x

The Attempt at a Solution

Okay, my terms at first would be u^n+2 +u^n. Then, du/dx = sec^2x, so i get du/1+u^2 = dx.

In the end i got (u^n+2 +u^n)/1+u^2. From then on I am stuck. Or did i do something wrong at first?

 

[Arkuski]

You know that u=tan[x] so du=sec²[x]dx.

Let's rearrange the integral a little bit to show the following:

∫tanⁿ[x](tan²[x]+1)dx

Remember that dx=du/sec²[x]. However, we need to find sec²[x] in terms of u. We use the following identity of sec²[x]=tan²+1 to show now that dx=du/(u²+1)

∫uⁿ(u²+1)/(u²+1)=∫uⁿ

This now becomes a simple integration problem and you should be able to do the rest.

 

[thoradicus]

oh great! didnt realize the numerator can be factorised like that. Thanks.