Problem 37 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d \theta}{1+\sin \theta}$$

Step-by-Step Solution

Answer

Question: Calculate the integral of the given function: \(\int \frac{d \theta}{1+\sin \theta}\) Answer: \(\int \frac{d\theta}{1+\sin\theta}=\ln\left|\frac{1+2\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\right|+C\)

Step 1: Double angle identity substitution

We can use the double angle identity for sine, which states that \(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\sin\theta\). Thus, we can rewrite the given integral as follows: $$\int \frac{d \theta}{1+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}$$

Step 2: Substitution of variable

Let \(u = \sin\frac{\theta}{2}\). Then \(\frac{d}{d\theta}\left(u\right)=\frac{1}{2}\cos\frac{\theta}{2}\). Therefore, we have: $$d\theta=2\frac{du}{\sqrt{1-u^2}}$$ Now substitute these expressions for \(\sin\frac{\theta}{2}\) and \(d\theta\) into the integral: $$\int \frac{2\frac{du}{\sqrt{1-u^2}}}{1+2u}$$

Step 3: Simplify Integral

Now, our integral becomes: $$\int \frac{2du}{(1+2u)\sqrt{1-u^2}}$$

Step 4: Perform Integration

Observe that if we perform the substitution \(v=u^2\), we can simplify the integral further. Performing this substitution yields: $$\int \frac{dv}{(1+2\sqrt{v})\sqrt{1-v}}$$ Now, perform the substitution \(w=1+2\sqrt{v}\) and simplify the integral: $$\int \frac{dw}{w(w-1)}$$ Finally, use partial fractions to decompose the integrand: $$\int \frac{dw}{w(w-1)}=\int\left(\frac{1}{w}+\frac{1}{w-1}\right)dw$$ Now, integrate each term separately: $$\int\frac{dw}{w}+\int\frac{dw}{w-1}=\ln|w|-\ln|w-1|+C$$

Step 5: Back substitution

Now substitute each variable as follows: $$= \ln\left|\frac{w}{w-1}\right|+C$$ $$= \ln\left|\frac{1+2\sqrt{v}}{1+2\sqrt{v}-1}\right|+C$$ $$= \ln\left|\frac{1+2\sqrt{v}}{2\sqrt{v}}\right|+C$$ $$= \ln\left|\frac{1+2\sqrt{u^2}}{2\sqrt{u^2}}\right|+C$$ $$= \ln\left|\frac{1+2\sqrt{\sin^2\frac{\theta}{2}}}{2\sqrt{\sin^2\frac{\theta}{2}}}\right|+C$$ Now, we have the final solution: $$\int \frac{d\theta}{1+\sin\theta}=\ln\left|\frac{1+2\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\right|+C$$