Question

Find the integrals of the function ${{\sin }^{2}}(2x+5)$.

Answer 1

Convert the square of sine function into cosine function of power one that is used $\cos 2x=1-2{{\sin }^{2}}x$.

Complete step by step answer:
Write the given function
$\int{{{\sin }^{2}}(2x+5)}dx$ …… (1)
Let $t=2x+5$ and differentiate. Therefore,

& dt=2dx+0 \\\ & dt=2dx \\\ & dx=\frac{dt}{2} \\\ \end{aligned}$$ Substitute the value of $$dx$$ in equation (1) and solve: $$\begin{aligned} & \int{{{\sin }^{2}}(2x+5)}dx=\int{{{\sin }^{2}}t}\frac{dt}{2} \\\ & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{2}\int{{{\sin }^{2}}t}dt \\\ \end{aligned}$$ …… (2) Since, $$\begin{aligned} & \cos 2t=1-2{{\sin }^{2}}t \\\ & 2{{\sin }^{2}}t=1-\cos 2t \\\ & {{\sin }^{2}}t=\frac{1-\cos 2t}{2} \\\ \end{aligned}$$ Now, substitute the value of $${{\sin }^{2}}t$$ in equation (2) and solve: $$\begin{aligned} & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{2}\int{\left( \frac{1-\cos 2t}{2} \right)}dt \\\ & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\int{\left( 1-\cos 2t \right)}dt \\\ & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\left[ \int{dt-\int{\cos 2t}}dt \right] \\\ & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\left[ t-\frac{\sin 2t}{2} \right]+C \\\ & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}t-\frac{\sin 2t}{8}+C \\\ \end{aligned}$$ …… (3) Where, C is the integration constant. Now, substitute back the value of t in equation (3) $$\begin{aligned} & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\left( 2x+5 \right)-\frac{\sin 2\left( 2x+5 \right)}{8}+C \\\ & \int{{{\sin }^{2}}(2x+5)}dx=\frac{\left( 2x+5 \right)}{4}-\frac{\sin \left( 4x+10 \right)}{8}+C \\\ \end{aligned}$$ **Note:** Integration of square of sine function is not possible so it is mandatory to convert the square of sine function into a function of cosine with power one.