Chapter 17: Problem 14

Find $\int \frac{\cos t-\sin t}{\sin t+\cos t} \mathrm{~d} t$.

Short Answer

Expert verified

Question: Evaluate the integral of the function $f(t) = \frac{\cos t - \sin t}{\sin t + \cos t}$. Answer: The integral of the function $f(t) = \frac{\cos t - \sin t}{\sin t + \cos t}$ is given by: $$\int \frac{\cos t-\sin t}{\sin t+\cos t} \mathrm{~d} t = \frac{1}{2} \ln\left| \frac{(\sin t - \cos t) - 1}{(\sin t - \cos t) + 1} \right| + (\sin t - \cos t)^2 + C$$ where $C$ is the constant of integration.

Step by step solution

Multiply and divide by the conjugate of the denominator

To simplify the given expression, multiply and divide the integrand by the conjugate of the denominator, which is $(\sin t - \cos t)$: $$ \int \frac{(\cos t - \sin t)(\sin t - \cos t)}{(\sin t + \cos t)(\sin t - \cos t)} \mathrm{~d} t $$

Simplify the numerator and denominator

We can simplify the numerator and denominator of the integrand by applying the difference of squares formula for conjugates. In our case, we will have: $$ \int \frac{(\cos^2 t - 2(\cos t)(\sin t) + \sin^2 t)}{(\sin^2 t - \cos^2 t)} \mathrm{~d} t $$

Simplify the integration further using trigonometric identities

Apply the identity $\cos^2 t + \sin^2 t = 1$ and simplify the expression inside the integral: $$ \int \frac{(1 - 2(\cos t)(\sin t))}{(\sin^2 t - \cos^2 t)} \mathrm{~d} t $$

Split the integral into two integrals

Now split the integral into two separate integrals: $$ \int \frac{1\mathrm{~d} t}{(\sin^2 t - \cos^2 t)} - \int \frac{2(\cos t)(\sin t)\mathrm{~d} t}{(\sin^2 t - \cos^2 t)} $$

Evaluate the first integral

To evaluate the first integral, we can observe that the integration has the form of $\frac{\mathrm{~d} t}{(a^2 -b^2)}$ and we can use the substitution method as follows: Let $u=\sin t - \cos t$, then $\mathrm{d}u = (\cos t + \sin t) \mathrm{~d} t$, so we have: $$ \int \frac{1\mathrm{~d} t}{(\sin^2 t - \cos^2 t)} = \int \frac{\mathrm{d}u}{u^2 - 1} $$ Now apply the hyperbolic secant antiderivative to the integral above: $$\int \frac{\mathrm{d}u}{u^2 - 1} = \frac{1}{2} \ln\left| \frac{u - 1}{u + 1} \right| + C_1 $$ Next, substitute the value of $u$ back into the solution: $$\frac{1}{2} \ln\left| \frac{(\sin t - \cos t) - 1}{(\sin t - \cos t) + 1} \right| + C_1 $$

Evaluate the second integral

To evaluate the second integral, we can use the same substitution method as in Step 5: Let $u=\sin t - \cos t$, then $\mathrm{d}u = (\cos t + \sin t) \mathrm{~d} t$, so we have: $$ \int \frac{2(\cos t)(\sin t)\mathrm{~d} t}{(\sin^2 t - \cos^2 t)} = \int 2u \mathrm{~d} u $$ Now integrate the expression: $$\int 2u \mathrm{~d} u = u^2 + C_2 $$ Next, substitute the value of $u$ back into the solution: $$ (\sin t - \cos t)^2 + C_2 $$

Combine the results and write the final answer

Now we can combine the results from Steps 5 and 6 to write the final answer for the integral: $$\frac{1}{2} \ln\left| \frac{(\sin t - \cos t) - 1}{(\sin t - \cos t) + 1} \right| + (\sin t - \cos t)^2 + C $$ Where $C = C_1 + C_2$ is the constant of integration. As a result, we have: $$\int \frac{\cos t-\sin t}{\sin t+\cos t} \mathrm{~d} t = \frac{1}{2} \ln\left| \frac{(\sin t - \cos t) - 1}{(\sin t - \cos t) + 1} \right| + (\sin t - \cos t)^2 + C $$

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Find \(\int \frac{\cos t-\sin t}{\sin t+\cos t} \mathrm{~d} t\).

Short Answer

Step by step solution

Multiply and divide by the conjugate of the denominator

Simplify the numerator and denominator

Simplify the integration further using trigonometric identities

Split the integral into two integrals

Evaluate the first integral

Evaluate the second integral

Combine the results and write the final answer

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