Integration Question (1 Viewer)

[echelon4]

(S=integral sign)

1. S dx/(4+5cosx)

2. S dx/(1+sin^2(x))

thanks in advanced

 

[A l]

Use t-formulae substitution for both of them...

 

[LoneShadow]

using t = tan[x/2]; sin[x] = 2t/(1+t^2); cos[x] = (1-t^2)/(1+t^2); dx/dt = 2/(1+t^2)

first one:

Integral[dx/{4+5cos(x)}] = Integral[2dt/(3^2-t^2)] = ln[(t+3)/(t-3)]/3 + C = ln[(tan{x/2}+3)/(tan{x/2}-3)]/3 +C


Use a similar method for the second one. It might be a bit messier, but it should work out at the end. You may need partial fractions.

 

[felixcthecat]

second one: S dx/(1+sin²x)

use double angle formula to change sin²x

should get S 2dx/(cos2x+1)

then use t substitutions, t=tanx

should be quite easy to integrate after this!

 

[LoneShadow]

second one: S dx/(1+sin^2(x))

use double angle formula to change sin^2(x)

should get S 2dx/(cos2x+1)

then use t substitutions, t=tanx

should be quite easy to integrate after this!

I liked your first answer better

 

[felixcthecat]

realli? aww and i took all the time to find how to type them hahaha.. umm


well obviously i like the edited version better orelse i wouldn't edit it.. ohwell~

i might as well also add an edit that "Let S = integral sign" lolz

 

[echelon4]

thanks for the replies.

man, i spent like an hour trying to work them out without using the t-formula.

 

[LoneShadow]

thanks for the replies.

man, i spent like an hour trying to work them out without using the t-formula.

It wasn't a waste. The higher you go, the longer you'll spend on doing a question.

It's a good idea to change your approach to the question when one fails.

 

[Raginsheep]

Just continue to practice and you should soon get a grip on when to use what.