Having trouble with a certain calculus integration technique?

Question

I checked out this calculus BC textbook from my school over the summer, just so I could do the problems and not forget what I learned in AB; anyway, I finished the AB sections, and am into BC. I am now seeing integrations as: integral(cos(x)^3sin(x)^5); the instructions on how to get through these problems are massive, and simply involve using specific changing of the cosine and sine functions, instead of using simple substitutions.

I don't have any teacher assistance; so, are there any internet users privy to this area of integrals. Please, I appreciate any help from any users.

Thank you.

Answer 1

sinx^5.cosx^3 = sinx^5(1 - sinx^2).cosx



u = sinx

du/dx = cosx



Then you get..... u^5(1 - u^2) du

hope it helped!

Answer 2

I consume each and every of the chocolate off the exterior, then I unroll and consume the cream by utilising scraping it off with my finger... then i'm getting unhappy because of the fact I ate each and every of the coolest areas first an now all I fairly have left is the effort-free old cake. Then i'm getting satisfied as quickly as I understand there is yet another one... then I do a similar ingredient as quickly as extra.

Answer 3

cos(x)^3 =>

cos(x)^2 * cos(x) =>

(1 - sin(x)^2) * cos(x) =>

cos(x) - sin(x)^2 * cos(x)



Multiply that by sin(x)^5 * dx



sin(x)^5 * cos(x) * dx - sin(x)^7 * cos(x) * dx



Now we use substitution



u = sin(x)

du = cos(x) * dx



Now we have:



u^5 * du - u^7 * du



Integrate



(1/6) * u^6 - (1/8) * u^8 + C =>

(1/24) * u^6 * (4 - 3u^2) + C



Back-substitute



(1/24) * sin(x)^6 * (4 - 3sin(x)^2) + C