Evaluate the integral of 5^t * sin (5^t) *dt

I started out with u = 5^t , but then I got stuck on du because I am not sure how to take the derivative of 5^t?

The answer from the book is
(-1/ln5) cos(5^t) + C

I understand the part with the antiderivative of sin being -cos, but that is about it.

You need to recall that

d/dt a^t = ln(a) a^t
because a = e^(lna), so

a^t = e^(lna)^t = e^(lna t)
so, using the chain rule,
d/dt e^(lna t) = lna e^(lna t) = lna a^t

u = 5^t
du = ln5 5^t

∫ 5^t sin(5^t) dt
= ∫1/ln5 sin(u) du
= 1/5 (-cos u)
= -1/5 cos(5^t) + C

Well, fear not, my friend! I'm here to shed some light on this integral conundrum, all while sprinkling a little humor into the mix.

First, let's address your question about the derivative of 5^t. It's true that taking the derivative of 5^t might initially seem like a tricky task. But don't worry, I'm about to reveal the magical secret behind it.

To find the derivative of 5^t, we can use a nifty little rule called the chain rule. The chain rule states that if we have a function raised to the power of another function, like in this case with 5^t, we simply take the derivative of the base (which is 5) and multiply it by the natural logarithm of the base (ln(5)), and then multiply the whole thing by the derivative of the exponent (which is just 1). So, the derivative of 5^t is ln(5) * 5^t.

Now that we've cleared that up, let's move on to the integral itself.

Since we're integrating with respect to t, we can rewrite the given integral as ∫(5^t * sin(5^t)) dt.

Using the marvelous power of substitution, you rightly set u = 5^t. And as a result, we have du = (ln(5) * 5^t) dt.

Now, if we simplify, the original integral becomes ∫(sin(u)) du.

The good news is we already know the antiderivative of sin(u) is -cos(u). So, we can go ahead and integrate to get -cos(u) + C, where C is the constant of integration.

Now, with our friendly substitution u = 5^t in mind, we replace u with 5^t to find the final answer of the integral:

∫(5^t * sin(5^t)) dt = -cos(5^t) + C.

But wait, there's a little extra step here! We need to divide our final answer by ln(5) because of the substitution with u earlier. So, β voilà! The correct answer, as stated in your book, is:

(-1/ln(5)) * cos(5^t) + C.

I hope this explanation brought some clarity, and hopefully, a smile to your face. Keep calm, and keep integrating with joy!

To evaluate the integral ∫5^t * sin(5^t) * dt, you correctly started by setting u = 5^t.

Now, let's find the derivative du/dt. To do this, we will use the chain rule.

Taking the derivative of both sides of u = 5^t with respect to t, we get:

du/dt = d/dt (5^t)

To find d/dt (5^t), we can use the exponential rule:

d/dt (a^t) = (ln a) * (a^t)

In this case, a = 5. So, substituting this into the equation, we have:

du/dt = (ln 5) * (5^t)

Now, notice that du/dt can also be written as (ln 5) * u. So, we can rewrite the integral as:

∫u * sin(u) * (1/(ln 5)) * du

Using the antiderivative of sin(u), which is -cos(u), we rewrite the integral as:

(1/(ln 5)) * ∫u * (-cos(u)) * du

Now, we can integrate this using the method of integration by parts. Let's use u = u and dv = -cos(u) * du. Then, we have du = (ln 5) * u * du and v = -sin(u).

Applying the integration by parts formula, we have:

∫u * (-cos(u)) * du = uv - ∫v * du

Substituting the values we found earlier, we get:

∫u * (-cos(u)) * du = (-u * sin(u)) - ∫(-sin(u)) * ((ln 5) * u) * du

Simplifying this, we have:

∫u * (-cos(u)) * du = -u * sin(u) + (ln 5) * ∫sin(u) * u * du

Now, notice that the remaining integral, ∫sin(u) * u * du, is exactly the same as the original integral we started with, but with u instead of t. Therefore, we can substitute back u = 5^t:

∫sin(u) * u * du = ∫sin(5^t) * (5^t) * dt

So, we have:

∫u * (-cos(u)) * du = -u * sin(u) + (ln 5) * ∫sin(u) * u * du
= -u * sin(u) + (ln 5) * ∫sin(5^t) * (5^t) * dt

This is a recursive equation, which means that it is essentially the same equation on both sides. We can solve this equation by solving for the original integral, since we have the value for ∫sin(u) * u * du:

∫sin(u) * u * du = (-1/(ln 5)) * u * cos(u) + C

Substituting this back into the equation, we get:

∫u * (-cos(u)) * du = -u * sin(u) + (ln 5) * [∫sin(5^t) * (5^t) * dt]
= -u * sin(u) + (ln 5) * [(-1/(ln 5)) * u * cos(u) + C]

Simplifying this further, we have:

∫5^t * sin(5^t) * dt = (-1/(ln 5)) * 5^t * cos(5^t) + C

So, the final answer is:

∫5^t * sin(5^t) * dt = (-1/(ln 5)) * cos(5^t) + C

To evaluate the integral ∫ 5^t * sin(5^t) dt, you correctly identified a substitution by setting u = 5^t. However, to continue, let's differentiate both sides of the equation u = 5^t with respect to t to find du/dt:

du/dt = d(5^t) / dt

To differentiate 5^t with respect to t, we can use the chain rule. We'll treat 5 as a constant and differentiate t as if it was any other variable, using the power rule:

du/dt = (ln 5) * (5^t)

Now that we have found du/dt, we multiply both sides by dt to isolate dt:

du = (ln 5) * (5^t) * dt

Substituting this result back into the original integral, we get:

∫ 5^t * sin(5^t) dt = ∫ sin(u) du

Using the antiderivative of sin(x) as -cos(x), we can evaluate the integral:

∫ sin(u) du = -cos(u) + C

Finally, substituting u = 5^t back into the result, we have:

∫ 5^t * sin(5^t) dt = -cos(5^t) + C

However, note that there is one more thing we need to take into account. Since we were working with the substitution u = 5^t, when substituting back, we also need to account for the derivative of u with respect to t. In this case, du/dt = (ln 5) * (5^t). So, we divide the result by du/dt:

∫ 5^t * sin(5^t) dt = (-1 / ln 5) * cos(5^t) + C

Thus, the correct answer is (∫ 5^t * sin(5^t) dt) = (-1 / ln 5) * cos(5^t) + C.