You are Here: Home

# P3 integration watch

1) integrate tan(x + Pi/4)

2) integrate xsec^2x

3) integrate x^2 e^x^3

I never know whether to do substitution or integration by parts

Thank you x
2. 1) Substitution of u = cos (x + pi/4)
2) Parts, u = x, dv/dx = sec^2 x
3) Substitution u = x^3
3. ok i can do number 3 now..still dont understand 1 and 2 though, could you explain a bit more please? xx
4. 1) int tan(x + pi/4) = int sin(x + pi/4)/cos(x + pi/4) Let u = cos(x + pi/4), du/dx = -sin(x + pi/4) so dx = -du/sin(x + pi/4)
Putting these bits into the integration, we get int sin(x + pi/4)/u * -du/sin(x + pi/4) = int -1/u du = -ln|u| + c = -ln|cos(x + pi/4)| + c = ln|sec(x + pi/4)| + c

2) int xsec^2 x Let u = x, dv/dx = sec^2 x so du/dx = 1, v = tanx
int xsec^2 x = xtanx - int tanx dx = xtanx + ln|cosx| + c
5. 1)

∫tan(x + pi/4)

= ∫sin(x + pi/4) / cos(x + pi/4)dx

let u = cos (x + pi/4)

so du/dx = -sin(x + pi/4)

so replace cos(x + pi/4) with u and dx by du/(-sin(x + pi/4))

-∫1/u du

= -ln(cos(x + pi/4)) + C

_______________________________

∫xsec²x dx

integration by parts:

u= x dv= sec²x

du=dx v= tanx

by definition the integral is

uv - ∫v(du)

= xtanx - ∫tanx dx

= xtanx + lncosx + C

Hope this helps.
6. (Original post by jazzyj)
ok i can do number 3 now..still dont understand 1 and 2 though, could you explain a bit more please? xx
2) INT x.(sec x)^2 dx

use parts, making u=x, dv/dx=(secx)^2
so, du/dx=1 and v=(tan x)
then just sub into the formula
7. thank you all so much!

by the way, is it just a standard result that you have learnt that the integral of tanx is ln cosx?
8. (Original post by jazzyj)
thank you all so much!

by the way, is it just a standard result that you have learnt that the integral of tanx is ln cosx?
The integral of tanx is ln|secx| = - ln|cosx| You can show it the same way as Q1 - substitution of u = cosx, du/dx = -sinx etc.

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: June 2, 2004
Today on TSR

### How do other unattractive guys cope?

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE