integrate (sin x)to power 6/ (cos x)?

Question

integrate (sin x)to power 6/ (cos x)

integrate (sin x)raised to 6/ (cos x)?

Answer 1

multiply and devide by cos x

(sin x)^6 cosx /(1-sin^2 x)

let sin x = t
cos x dx = dt
we have to integrate t^6/(1-t^2) dt

= (t^6 -1+1)/(1-t^2)
or -(1+t^2+t4) + 1/(1-t^2)

- (1+t^2+t4) + (1/(1-t) - 1/(1+t) )/2

now each element can be integrated

Answer 2

Rather than actually doing the problem, I think it would be more helpful to you in the long run to point you to a couple of sources for trigonometric integration:

http://www.physicsforums.com/showthread.php?threadid=138297

http://people.hofstra.edu/faculty/stefan_waner/realworld/trig/trig4.html

Hint: If this problem were to be given to me (as one similar probably was a couple of years ago when I took calc II), my first instinct would be to integrate by parts, which is addressed in the first link, but you might find an easier way by examining different trig identities (there are a lot of them out there)...
Hope this helps!

Answer 3

using fact the spinoff of tan(x) is sec²(x), ?tan(x) sec²(x) dx = (one million/2)tan²(x) + C ---- replace: the rationalization which you're starting to be a super form of distinctive looking solutions below is as a results of the fact one million+tan²(x) = sec²(x). in particular, think that C = one million/2 + D Then the respond could be (one million/2)tan²(x) + one million/2 + D = (one million/2)(tan²(x) + one million) + D = (one million/2)sec²(x) + D = (one million/2)(one million/cos²(x)) + D those solutions are all equivalent. it extremely is between the few circumstances the place that consistent you mostly tack on after integration impacts what the effect sounds like. So usually, to income if 2 integrations of a similar function are equivalent, you ought to subtract the two effects and if the version is a persevering with, you be attentive to that the two solutions are merely as valid (or merely as incorrect, for that count).

Answer 4

∫ [sin(x)]^6 / cos(x) dx

= ∫ [sin^2(x)]^3 / cos(x) dx

= ∫ [1 - cos^2(x)]^3 / cos(x) dx, because sin^2(x) + cos^2(x) = 1

Expand the numerator.

= ∫ [1 - 3cos^2(x) + 3cos^4(x) - cos^6(x)] / cos(x) dx

Divide each numerator term by cos(x).

= ∫ [1 / cos(x) - 3cos(x) + 3cos^3(x) - cos^5(x)] dx

Separate the terms.

={ ∫1 / cos(x) dx} - {3∫cos(x) dx} + {3∫cos^3(x) dx} - {∫cos^5(x) dx}

After looking up the difficult ones in a manual :

= {ln[tan(x/2 + pi/2)]} - {3sin(x)} + {3sin(x) - sin^3(x)} - {13sin(x) / 5 - 8sin^3(x) / 5 + 4sin^5(x) / 5}

Simplifying.

= ln[tan(x/2 + pi/2)] - [sin(x)][13 + 4sin^4(x) - 3sin^2(x)] / 5

I hope that's correct.

Answer 5

If you tell me what the meaning of power....i'll help you..i'm from spain

Answer 6

very difficult