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Integration by substitution

I'm rubbish at integration by substitution and I just can't get it right! Plus I can't figure out how I've gone wrong either, if someone can figure out where I've gone wrong with this question please (I'll post my working from my phone now)

(Integrate) 1/(cos^2x)(1+2tanx)^2 when u=1+2tanx

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Reply 1

mica-lwe

ImageUploadedByStudent Room1422310893.532712.jpg

question 8

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Reply 2

TeeEm

Original post by mica-lwe

question 8

Posted from TSR Mobile

1/u² does not integrate to lnu

Reply 3

jamez870

Instead of writing what dx is equal to, write it as what du is equal to

Reply 4

mica-lwe

Original post by TeeEm

1/u² does not integrate to lnu

Is that where I've gone wrong then? Is it better to take 1/2 out the integral and integrate u^-2?

Reply 5

TeeEm

Original post by mica-lwe

Is that where I've gone wrong then? Is it better to take 1/2 out the integral and integrate u^-2?

sure

then you are one step from the required answer

Reply 6

mica-lwe

Original post by TeeEm

sure

then you are one step from the required answer

thankyou! Have no idea why I struggle with substitution when there's harder stuff on the paper :colondollar:

Reply 7

TeeEm

Original post by mica-lwe

thankyou! Have no idea why I struggle with substitution when there's harder stuff on the paper :colondollar:

practice
if this fails try practice
and as a back up technique try practice...

http://www.madasmaths.com/archive/maths_booklets/further_topics/integration/integration_definite_mixed_up.pdf

http://www.madasmaths.com/archive/maths_booklets/standard_topics/integration/integration_structured_exam_questions_part_i.pdf

Reply 8

mica-lwe

Original post by TeeEm

I'll try these, thanks! :biggrin:

Reply 9

TeeEm

Original post by mica-lwe

I'll try these, thanks! :biggrin:

long download time but full solutions

all the best

Reply 10

Bob2222

hi,

Your solution makes a good start... it is fine until the line =1/2 ∫ 1/(u^2) du

keep the half outside the integral sign to simplify things; then 1/(u^2) is the same as u^(-2) because the reciprocal of a number makes the power negative and:

∫ u^(-2) du = -1/u +c as you divide by the power then add one to it ...don't forget the constant (c).

so... 1/2 ∫ u^(-2) du = -1/2u +c and you can just swap u for the substitution you made at the start.

Hope that helps :smile: