Question

The integral of cos^5 tdt

Help!?!?

Answer 1

First write
\[ \cos^5 = \cos (1 - \sin^2)^2 \]

Now you can substitute \( u = \sin x \) and it becomes fairly easy

Answer 2

Can you solve it..Because Im totally black. :/

Answer 3

\[\large \int\limits\limits \cos^5t \;dt \qquad = \qquad \int\limits \cos t \;\cos^2t \;\cos^2 t \;dt\]Using this identity,\[\large \color{salmon}{\cos^2t+\sin^2t=1}\qquad\rightarrow\qquad \color{salmon}{\cos^2t=1-\sin^2t}\]
\[\int\limits\limits \cos t \;\color{salmon}{\cos^2t \;\cos^2 t} \;dt \qquad \rightarrow \qquad \int\limits \cos t\;\color{salmon}{(1-\sin^2t)(1-\sin^2t)}dt\]

Which as James showed, we can write as,\[\large \int\limits \cos t(1-\sin^2t)^2 dt\]

Answer 4

Let,\[\large \color{orangered}{u=\sin t}\]Taking the derivative gives us,\[\large \color{royalblue}{du=\cos t \;dt}\]

Now plug those pieces into here to make the substitution.
\[\large \int\limits\limits (1-\color{orangered}{\sin}^2\color{orangered}{t})^2 \color{royalblue}{\cos t\;dt}\]

Answer 5

Giving us,\[\large \int\limits (1-\color{orangered}{u}^2)^2 \; \color{royalblue}{du}\]

Answer 6

We'll expand out the brackets.\[\large \int\limits 1-2u^2+u^4 \; du\]

Integrating, applying the Power Rule for Integration to each term, gives us,\[\large u-\frac{2}{3}u^3+\frac{1}{5}u^5+C\]

Our final step will be to back-substitute.
We need it back in terms of `t` since that's what we started with.

Answer 7

\[\large \color{orangered}{u}-\frac{2}{3}\color{orangered}{u}^3+\frac{1}{5}\color{orangered}{u}^5+C \qquad \rightarrow \qquad \color{orangered}{\sin t}-\frac{2}{3}\color{orangered}{\sin}^3\color{orangered}{t}+\frac{1}{5}\color{orangered}{\sin}^5\color{orangered}{t}+C\]

Answer 8

So there ya go.
Lemme know if you're confused on any of those steps.