Question

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

Answer 1

Hint: 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,
\[\begin{align}
  & dt=2dx+0 \\
 & dt=2dx \\
 & dx=\frac{dt}{2} \\
\end{align}\]
Substitute the value of \[dx\] in equation (1) and solve:
\[\begin{align}
  & \int{{{\sin }^{2}}(2x+5)}dx=\int{{{\sin }^{2}}t}\frac{dt}{2} \\
 & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{2}\int{{{\sin }^{2}}t}dt \\
\end{align}\] …… (2)
Since,
 \[\begin{align}
  & \cos 2t=1-2{{\sin }^{2}}t \\
 & 2{{\sin }^{2}}t=1-\cos 2t \\
 & {{\sin }^{2}}t=\frac{1-\cos 2t}{2} \\
\end{align}\]
Now, substitute the value of \[{{\sin }^{2}}t\] in equation (2) and solve:
\[\begin{align}
  & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{2}\int{\left( \frac{1-\cos 2t}{2} \right)}dt \\
 & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\int{\left( 1-\cos 2t \right)}dt \\
 & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\left[ \int{dt-\int{\cos 2t}}dt \right] \\
 & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\left[ t-\frac{\sin 2t}{2} \right]+C \\
 & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}t-\frac{\sin 2t}{8}+C \\
\end{align}\] …… (3)
Where, C is the integration constant.
Now, substitute back the value of t in equation (3)
\[\begin{align}
  & \int{{{\sin }^{2}}(2x+5)}dx=\frac{1}{4}\left( 2x+5 \right)-\frac{\sin 2\left( 2x+5 \right)}{8}+C \\
 & \int{{{\sin }^{2}}(2x+5)}dx=\frac{\left( 2x+5 \right)}{4}-\frac{\sin \left( 4x+10 \right)}{8}+C \\
\end{align}\]

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.