How do you find the integral of $[{(sin t)}^{7}] [{(cos t)}^{4}] d t$ $[(sin t)^7][(cos t)^4]dt$ ?

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How do you find the integral of $[{(sin t)}^{7}] [{(cos t)}^{4}] d t$ $[(sin t)^7][(cos t)^4]dt$ ?

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Answer 1 · 2015-04-11T13:31:51

$\int [{(sin t)}^{7}] [{(cos t)}^{4}] d t$ $int [(sin t)^7][(cos t)^4]dt$

We know that $\frac{d}{d x} (sin x) = cos x$ $d/(dx)(sinx)=cosx$ and $\frac{d}{d x} (cos x) = - sin x$ $d/(dx)(cosx)= - sinx$ .

We'd like to do a substitution. We also know that :
${(sin t)}^{2} + {(cos t)}^{2} = 1$ $(sint)^2+(cost)^2=1$

Those are the key ideas.
We have $sin t$ $sint$ raised to an odd power and $cos t$ $cost$ raised to a power, so if we group one $sin t$ $sint$ witlh $d t$ $dt$ and change the remaining even power of $sin t$ $sint$ to an expression involving only $cos t$ $cost$ , we will be able to substitute.

$\int [{(sin t)}^{7}] [{(cos t)}^{4}] d t = \int {[{(sin t)}^{2}]}^{3} [{(cos t)}^{4}] [sin t] d t$ $int [(sin t)^7][(cos t)^4]dt = int [(sin t)^2]^3[(cos t)^4][sint]dt$

$= \int {[1 - {(cos t)}^{2}]}^{3} [{(cos t)}^{4}] [sin t] d t$ $= int [1-(cos t)^2]^3 [(cos t)^4][sint]dt$

Let $u = cos t$ $u=cost$ , then the integral becomes:

$= - \int {[1 - u^{2}]}^{3} [u^{4}] d u$ $= - int [1-u^2]^3 [u^4]du$

Expand the polynomial and integrate, then back-substitute for $u$ $u$ back to $cos t$ $cost$ .

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How do you find the integral of $[{(sin t)}^{7}] [{(cos t)}^{4}] d t$ $[(sin t)^7][(cos t)^4]dt$ ?

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How do you find the integral of $[{(sin t)}^{7}] [{(cos t)}^{4}] d t$ $[(sin t)^7][(cos t)^4]dt$ ?