Chapter 17: Problem 7

Find $\int \sin m t \sin n t \mathrm{~d} t$ where $m$ and $n$ are constants and $m \neq n .$

Short Answer

Expert verified

Question: Evaluate the integral of the product of two sine functions with different constants: $\int \sin(mt) \sin(nt) dt$, where m ≠ n. Answer: $\int \sin(mt) \sin(nt) dt = \frac{1}{2} \left[\frac{1}{m-n}\sin(mt-nt) - \frac{1}{m+n}\sin(mt+nt)\right] + C$

Step by step solution

Apply product-to-sum trigonometric identity

We apply the product-to-sum trigonometric identity: $\sin A \sin B = \frac{1}{2} (\cos(A-B) - \cos(A+B))$. Here, $A = mt$ and $B = nt$. So, $\sin(mt) \sin(nt) = \frac{1}{2} (\cos(mt-nt) - \cos(mt+nt))$. Now our integral becomes: $$\int \sin(mt) \sin(nt) dt = \int \frac{1}{2} (\cos(mt-nt) - \cos(mt+nt)) dt$$

Separate the integrals

Now we split the integral into two separate integrals: $$\int \frac{1}{2} (\cos(mt-nt) - \cos(mt+nt)) dt = \frac{1}{2} \int \cos(mt-nt) dt - \frac{1}{2} \int \cos(mt+nt) dt$$

Evaluate the integrals

Now, let's evaluate each integral individually: 1) $\int \cos(mt-nt) dt$: Substitute $u = mt-nt$, then $du = (m-n) dt$. Therefore, $dt = \frac{1}{m-n} du$. $$\int \cos(mt-nt) dt = \int \cos(u) \frac{1}{m-n} du = \frac{1}{m-n} \int \cos(u) du = \frac{1}{m-n}[\sin(u) + C_1]$$ 2) $\int \cos(mt+nt) dt$: Substitute $v = mt+nt$, then $dv = (m+n) dt$. Therefore, $dt = \frac{1}{m+n} dv$. $$\int \cos(mt+nt) dt = \int \cos(v) \frac{1}{m+n} dv = \frac{1}{m+n} \int \cos(v) dv = \frac{1}{m+n}[\sin(v) + C_2]$$

Combine the results

Now that we have evaluated both integrals, we will combine them and substitute back the original variables: $$\frac{1}{2} \int \cos(mt-nt) dt - \frac{1}{2} \int \cos(mt+nt) dt = \frac{1}{2} \left[\frac{1}{m-n}\sin(mt-nt) - \frac{1}{m+n}\sin(mt+nt)\right] + C$$ So, the final answer is: $$\int \sin(mt) \sin(nt) dt = \frac{1}{2} \left[\frac{1}{m-n}\sin(mt-nt) - \frac{1}{m+n}\sin(mt+nt)\right] + C$$

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Find \(\int \sin m t \sin n t \mathrm{~d} t\) where \(m\) and \(n\) are constants and \(m \neq n .\)

Short Answer

Step by step solution

Apply product-to-sum trigonometric identity

Separate the integrals

Evaluate the integrals

Combine the results

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