Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Integration by substituition

Question: Use the substitution x = ln u to find

\displaystyle\int \dfrac{e^x}{1+ e^{2x}}\ dx

I'm stuck in the end; here's how I tried to do it:

\dfrac{dx}{du} = \dfrac{1}{u}

so

dx = \dfrac{du}{u}

\Rightarrow

\displaystyle\int \dfrac{e^{\ln u}}{1+ e^{2\ln u}}\ \dfrac{du}{u}

\Rightarrow

\displaystyle\int \dfrac{1}{1+ u^2}\ du

I don't know how to do it after that last step . . .

(edited 12 years ago)

you have the numerator wrong in your last line

OP

Original post by TenOfThem

you have the numerator wrong in your last line

Original post by Fing4

e^(lnu) is not 1

Please see my post carefully, the numerator u is cancelled by 1/u when we replace dx by du/u.

Original post by Zishi

Please see my post carefully, the numerator u is cancelled by 1/u when we replace dx by du/u.

ooops

reading whilst eating :frown:

:frown:

In that case .... use the standard integral for

\frac{1}{1+x^2}

(edited 12 years ago)

Original post by Zishi

Question: Use the substitution x = ln u to find

\displaystyle\int \dfrac{e^x}{1+ e^{2x}}\ dx

I'm stuck in the end; here's how I tried to do it:

\dfrac{dx}{du} = \dfrac{1}{u}

so

dx = \dfrac{du}{u}

\Rightarrow

\displaystyle\int \dfrac{e^{\ln u}}{1+ e^{2\ln u}}\ \dfrac{du}{u}

\Rightarrow

\displaystyle\int \dfrac{1}{1+ u^2}\ du

I don't know how to do it after that last step . . .

Your working is correct so far, ignore the above posts.

Do you know the integrals/derivatives of the inverse trig funtions? You should be able to find one which matches this integral exactly.

OP

Original post by EEngWillow

Your working is correct so far, ignore the above posts.

Do you know the integrals/derivatives of the inverse trig funtions? You should be able to find one which matches this integral exactly.

Nope, but I guess I can workout derivatives/integrals of their inverse by working out from scratch. To solve this question do I need to learn some of 'em? :s-smilie:

:s-smilie:

OP

Original post by TenOfThem

In that case .... use the standard integral for

\frac{1}{1+x^2}

What's the standard integral?

tan^{-1}x

Original post by Zishi

Nope, but I guess I can workout derivatives/integrals of their inverse by working out from scratch. To solve this question do I need to learn some of 'em? :s-smilie:

:s-smilie:

Pretty much. I think they're in one of the FP modules?

Anyway,

\dfrac{d}{dx}arcsin(x) = \dfrac{1}{\sqrt{1-x^2}}

\dfrac{d}{dx}arccos(x) = -\dfrac{1}{\sqrt{1-x^2}}

\dfrac{d}{dx}arctan(x) = \dfrac{1}{x^2+1}

(edited 12 years ago)

As EEngWillow said ... you need to know the differentials of the inverse trig functions

http://www.cliffsnotes.com/study_guide/Differentiation-of-Inverse-Trigonometric-Functions.topicArticleId-39909,articleId-39884.html

I think you only use tan and sin in C4

(edited 12 years ago)

OP

Original post by EEngWillow

Pretty much. I think they're in one of the FP modules?

Anyway,

\dfrac{d}{dx}arcsin(x) = \dfrac{1}{\sqrt{1-x^2}}

\dfrac{d}{dx}arccos(x) = -\dfrac{1}{\sqrt{1-x^2}}

\dfrac{d}{dx}arctan(x) = \dfrac{1}{x^2+1}

Original post by TenOfThem

As EEngWillow said ... you need to know the differentials of the inverse trig functions

http://www.cliffsnotes.com/study_guide/Differentiation-of-Inverse-Trigonometric-Functions.topicArticleId-39909,articleId-39884.html

I think you only use tan and sin in C4

Thanks.

:smile:

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