An integration question: so the original problem was integral of (sinx)^3/ (cosx)^4, and I simplified this to integral of secx(tanx)^3, which is equal to int. of secx((secx)^2-1). This is where I am stuck. Any help would be much appreciated :)

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

An integration question: so the original problem was integral of (sinx)^3/ (cosx)^4, and I simplified this to integral of secx(tanx)^3, which is equal to int. of secx((secx)^2-1). This is where I am stuck. Any help would be much appreciated :)

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

How about instead of going the secant route...and I'm just tossing out a ball here rewrite to \[\int\limits_{?}^{?}(\sin x)^3 * 1/(\cos x )^4 = (\sin x )^2 * 1/(\cos x)^4 * (\sin x)\] then use the identity sin^2 x + cos^2 x = 1 to replace the first sin^2 x to get \[\int\limits_{?}^{?}1- (\cos x)^2 * 1/ (\cos x)^4 * \sin x\] The cos x's cancel and you get 1/(cos)^2 * sin x and then I'm not sure is that getting simpler? reuse the identity cos^2 = -sin^2 +1 which will give you \[\int\limits_{?}^{?}1/(\sin x)^2 +1 * (\sin x) / 1\]
nope now I'm stuck too...
\[\frac{\sin^{3} x}{\cos^{4} x} = \frac{\sin x(1-\cos^{2} x)}{\cos^{4} x}\] Let u = cos x du = -sin x \[\int\limits_{}^{}\frac{-(1-u^{2})}{u^{4}} = \int\limits_{}^{}\frac{u^{2}-1}{u^{4}} = \int\limits_{}^{}\frac{1}{u^{2}}-\frac{1}{u^{4}}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Ahhh I was On the right track...I was look...but then I tried to keep being fancy instead of going for the u sub.
so I kept my first simplification: \[\int\limits_{?}^{?} \sec x \tan ^{3} x\] then used u = sec x du = sec x tan x for \[\int\limits_{?}^{?} (u ^{2}-1)du = 1/3 (u ^{3}) + u +C\] but thanks for the help

Not the answer you are looking for?

Search for more explanations.

Ask your own question