Evaluate the integral by making the given substitution. (Use C for the constant of integration.)?

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Answer 1 · 2017-11-15T04:12:15

$\frac{{sin}^{6} (θ)}{6} + c$ $sin^6(theta)/6+c$

$\int {sin}^{5} (θ) cos (θ) d (θ)$ $intsin^5(theta)cos(theta)d(theta)$

with $u = sin (θ)$ $u=sin(theta)$

then the derivative is
$d u = cos (θ) d (θ)$ $du=cos(theta)d(theta)$

$d (θ) = \frac{1}{cos (θ)}$ $d(theta)=1/cos(theta)$
now in the integral

$\int {(u)}^{5} cos (θ) \frac{1}{cos (θ)} d (θ)$ $int(u)^5cos(theta)1/cos(theta)d(theta)$

the cos is goes

$\int {(u)}^{5} d (θ)$ $int(u)^5d(theta)$

integrating

$\int {(y)}^{n} d y = \frac{y^{n + 1}}{n}$ $int(y)^ndy=y^(n+1)/n$

$\int {(u)}^{5} d (θ) = \frac{u^{6}}{6} + c$ $int(u)^5d(theta)=u^6/6+c$

but $u = sin (θ)$ $u=sin(theta)$

$\int {sin}^{5} (θ) cos (θ) d (θ) = \frac{{sin}^{6} (θ)}{6} + c$ $intsin^5(theta)cos(theta)d(theta)=sin^6(theta)/6+c$

Answer 2 · 2017-11-15T04:13:49

$\frac{{sin}^{6} (θ)}{6} + C$ $sin^6(theta)/6 + C$

Using U-Substitution:
$u = sin (θ)$ $u = sin(theta)$
$d u = cos (θ)$ $du = cos(theta)$

Change the variables in your integral:
$\int {sin}^{5} (θ) cos (θ) d θ$ $int sin^5(theta)cos(theta)d theta$

$\int u^{5} d u$ $int u^5du$

Integrate:
$\int u^{5} d u =$ $int u^5du=$

$= \frac{u^{6}}{6} + C$ $=u^6/6 + C$

Replace u with $sin (θ)$ $sin(theta)$ :

$\frac{{sin}^{6} (θ)}{6} + C$ $sin^6(theta)/6 + C$