Evaluate the integral using the given substitution.
-$$\int \left( 4 - \sin \frac { t } { 5 } \right) ^ { 2 } \cos \frac { t } { 5 } d t , \quad u = 4 - \sin \frac { t } { 5 }$$
A) $\frac { 1 } { 3 } \left( 4 - \sin \frac { t } { 5 } \right) ^ { 3 } \sin \frac { t } { 5 } + C$
B) $5 \left( 4 - \sin \frac { t } { 5 } \right) ^ { 3 } + C$
C) $\frac { 5 } { 3 } \left( 4 - \cos \frac { t } { 5 } \right) ^ { 3 } + C$
D) $- \frac { 5 } { 3 } \left( 4 - \sin \frac { t } { 5 } \right) ^ { 3 } + C$

Question

Quizplus · Accepted Answer

Given $u = 4 - \sin \frac{t}{5}$, we differentiate both sides with respect to $t$ to find $du$. This gives $du = -\frac{1}{5} \cos \frac{t}{5} dt$, or equivalently, $dt = -5 \cos \frac{t}{5} du$. Substituting $u$ and $dt$ into the integral transforms it into $\int u^2 (-5) du$, which integrates to $-\frac{5}{3}u^3 + C$. Substituting $u$ back in terms of $t$ yields $-\frac{5}{3}(4 - \sin \frac{t}{5})^3 + C$.

Evaluate the Integral Using the Given Substitution ∫(4−sin⁡t5)2cos⁡t5dt,u=4−sin⁡t5\int \left( 4 - \sin \frac { t } { 5 } \right) ^ { 2 } \cos \frac { t } { 5 } d t , \quad u = 4 - \sin \frac { t } { 5 }∫(4−sin5t​)2cos5t​dt,u=4−sin5t​

Evaluate the Integral Using the Given Substitution $\int \left( 4 - \sin \frac { t } { 5 } \right) ^ { 2 } \cos \frac { t } { 5 } d t , \quad u = 4 - \sin \frac { t } { 5 }$