Integration (1 Viewer)

[cutemouse]

How would I do this? Thanks



Thanks

 

[gurmies]

Sorry, I can't do this one - along with these two:

 

[MC Squidge]

where did u see these integrals??

 

[jet]

Just tell me of any errors, apart from missing the + C.

 

[jaztheman77]

, simplify
sin^2 x to 1/2 (1-cos2x)
cos ^2 x to 1/2(1+ cos 2x)

Then becomes easier

 

[vds700]

think u made a mistake. If you divided by cos²x, wouldnt you get (sec²x + tan²x)/(sec²x + 1)?

 

[study-freak]

Sorry, I can't do this one - along with these two:

I know how to do the last one but I don't have time since I have to go somewhere now.
So I'll type it up at night.

 

[azureus88]

[maths]let\,\,u^2=e^x\\2udu=e^xdx\\dx=\frac{2udu}{e^x}=\frac{2udu}{u^2}=\frac{2du}{u}\\\\I=\int \frac{1}{\sqrt{1-e^x}}dx\\=2\int \frac{du}{u\sqrt{1-u^2}}\\\\let u=\sin\theta\\du=\cos\theta d\theta\\\\I=2\int \frac{\cos\theta d\theta}{\sin\theta \cos\theta}\\=2\int \csc\theta d\theta\\=2\int \frac{\csc\theta (\csc\theta +\cot\theta)d\theta}{\csc\theta +\cot\theta}\\=-2\ln(\csc\theta +\cot\theta)\\=-2\ln(\frac{1}{u}+\frac{\sqrt{1-u^2}}{u})\\=-2\ln(\frac{1}{\sqrt{e^x}}+\frac{{\sqrt{1-e^x}}}{\sqrt{e^x}})+c[/maths]

sorry study-freak, didn't see your post until after i wrote it up

 

[jet]

think u made a mistake. If you divided by cos²x, wouldnt you get (sec²x + tan²x)/(sec²x + 1)?

Exactly. Thanks for pointing that out.

 

[azureus88]

For the first one, i guess you could use t-substitution. cant think of a better method at the moment.

[maths]let\,\,t=\tan x\\dt=\sec^2xdx=(1+t^2)dx\\dx=\frac{dt}{1+t^2}\\\\\int \frac{1+\sin^2x}{1+\cos^2x}dx\\=\int \left ( \frac{1+\frac{t^2}{1+t^2}}{1+\frac{1}{1+t^2}} \right ).\frac{dt}{1+t^2}\\=\int \frac{2t^2+1}{(t^2+2)(t^2+1)}dt\\$Then do partial fractions$[/maths]

 

[kwabon]

For the first one, i guess you could use t-substitution. cant think of a better method at the moment.

[maths]let\,\,t=\tan x\\dt=\sec^2xdx=(1+t^2)dx\\dx=\frac{dt}{1+t^2}\\\\\int \frac{1+\sin^2x}{1+\cos^2x}dx\\=\int \left ( \frac{1+\frac{t^2}{1+t^2}}{1+\frac{1}{1+t^2}} \right ).\frac{dt}{1+t^2}\\=\int \frac{2t^2+1}{(t^2+2)(t^2+1)}dt\\$Then do partial fractions$[/maths]

nah would take too long, you're previous method was good though.

 

[gurmies]

Thanks guys, I just did them earlier - weren't as hard as I excpected.

 

[study-freak]

[maths]let\,\,u^2=e^x\\2udu=e^xdx\\dx=\frac{2udu}{e^x}=\frac{2udu}{u^2}=\frac{2du}{u}\\\\I=\int \frac{1}{\sqrt{1-e^x}}dx\\=2\int \frac{du}{u\sqrt{1-u^2}}\\\\let u=\sin\theta\\du=\cos\theta d\theta\\\\I=2\int \frac{\cos\theta d\theta}{\sin\theta \cos\theta}\\=2\int \csc\theta d\theta\\=2\int \frac{\csc\theta (\csc\theta +\cot\theta)d\theta}{\csc\theta +\cot\theta}\\=-2\ln(\csc\theta +\cot\theta)\\=-2\ln(\frac{1}{u}+\frac{\sqrt{1-u^2}}{u})\\=-2\ln(\frac{1}{\sqrt{e^x}}+\frac{{\sqrt{1-e^x}}}{\sqrt{e^x}})+c[/maths]

sorry study-freak, didn't see your post until after i wrote it up

lol that's alrite..