Evaluate the integral using the given substitution.

Ask by Martinez Johnson. in Malaysia

Mar 27,2025

Question

Evaluate the integral using the given substitution. $$ \int\left(4-\sin \frac{t}{5}\right)^{2} \cos \frac{t}{5} d t, u=4-\sin \frac{t}{5} $$

UpStudy AI · Accepted Answer

We are given
$$
\int\left(4-\sin \frac{t}{5}\right)^{2} \cos \frac{t}{5}\, dt
$$
with the substitution
$$
u = 4-\sin \frac{t}{5}\,.
$$

**Step 1. Compute the derivative of $$ u $$ with respect to $$ t $$:**

Differentiate with respect to $$ t $$:
$$
\frac{du}{dt} = -\frac{1}{5}\cos \frac{t}{5}\,.
$$
This implies
$$
du = -\frac{1}{5}\cos \frac{t}{5}\, dt\,.
$$

**Step 2. Solve for $$ \cos \frac{t}{5}\, dt $$:**

Multiplying both sides by $$ -5 $$, we obtain:
$$
-5\, du = \cos \frac{t}{5}\, dt\,.
$$

**Step 3. Substitute into the integral:**

Replace $$ \left(4-\sin \frac{t}{5}\right)^{2} $$ by $$ u^2 $$ and $$ \cos \frac{t}{5}\, dt $$ by $$ -5\, du $$:
$$
\int u^2 \cos \frac{t}{5}\, dt = \int u^2 (-5\, du) = -5 \int u^2\, du\,.
$$

**Step 4. Evaluate the integral with respect to $$ u $$:**

Integrate:
$$
-5 \int u^2\, du = -5 \left(\frac{u^3}{3}\right) = -\frac{5}{3} u^3 + C\,.
$$

**Step 5. Substitute back in terms of $$ t $$:**

Recall that $$ u = 4-\sin \frac{t}{5} $$. Therefore, the final answer is:
$$
-\frac{5}{3}\left(4-\sin \frac{t}{5}\right)^3 + C\,.
$$
The integral evaluates to $$ -\frac{5}{3}\left(4-\sin \frac{t}{5}\right)^3 + C $$.

Evaluate the integral using the given substitution.

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