I want to know haw to resolve this integral

by **dbalinov** » Wed Jun 04, 2008 11:08 am

I want to know haw to resolve this integral!

∫tan²x dx

by **Math Tutor** » Thu Jun 05, 2008 7:26 am

[tex]\int tg^2xdx = \int \frac{sin^2x}{cos^2x }dx =[/tex]

[tex]\int \frac{1-cos^2x}{cos^2x } dx = \int \frac{1}{cos^2x }dx - \int 1 dx = tgx - x[/tex]

by **garion** » Thu Jun 05, 2008 8:11 am

teacher wrote:[tex]\int tg^2xdx = \int \frac{sin^2x}{cos^2x }dx =[/tex]

[tex]\int \frac{1-cos^2x}{cos^2x } dx = \int \frac{1}{cos^2x }dx - \int 1 dx = tgx - x[/tex]

[tex]\int \frac{1}{cos^2x }dx - \int 1 dx = tgx - x + c[/tex]
where c=const

by **Math Tutor** » Thu Jun 05, 2008 11:59 am

Thank you garion!
Absolutely true

I want to know haw to resolve this integral

I want to know haw to resolve this integral

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