Question

$$ int ( sin \theta + cos \theta ) ^ { 2 } d \theta $$

Explanation:

Step1: Expand the integrand

$$\begin{align*} (\sin\theta+\cos\theta)^2&=\sin^{2}\theta + 2\sin\theta\cos\theta+\cos^{2}\theta\\ &=1 + \sin(2\theta) \end{align*}$$

Step2: Integrate term - by - term

$$\begin{align*} \int(\sin\theta+\cos\theta)^2d\theta&=\int(1 + \sin(2\theta))d\theta\\ &=\int 1d\theta+\frac{1}{2}\int\sin(2\theta)d(2\theta) \end{align*}$$

Step3: Evaluate the integrals

$$\begin{align*} \int 1d\theta&=\theta+C_1\\ \frac{1}{2}\int\sin(2\theta)d(2\theta)&=-\frac{1}{2}\cos(2\theta)+C_2 \end{align*}$$

Answer:

$\theta-\frac{1}{2}\cos(2\theta)+C$ ($C = C_1 + C_2$ is the constant of integration)