∫sin^2x/cos^4x dx

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∫sin^2x/cos^4x dx

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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= tan^2 x sec^2x
sin^2 + cos^2 =1 tan^2 + 1 = sec^2
so integrad becomes sec^2(x) [ sec^2 (x) -1 ]

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Other answers:

expanding sec^4(x) -sec^2(x) , now sec^2 integrals to tan so we look to rewrite the sec^4(x)
Thank you but how can i rewrite sec^4(x)
yeh , by checking with wolframa "http://www.wolframalpha.com/input/?i=xtan%28x%29" there is no way to do this with only elementary functions
your teacher shouldnt be asking you to do this question
if a software package has trouble doing it then how can they expect a student to do it
This question was in my book the answer is 1/3tan^3(x) but hhats all they've wrote
yeh im so stupid
when you get to tan^2(x)sec^2(x) you do a substittuion, u= tan(x) so du= sec^2(x) dx
so you get the integral of u^2 du , which gives the answer
Thank you!

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