Question

Please help me with my calculus homeworkI am paying close attentionThanks a lot

Answer 1

`(1)/(8)\sin ^4(2x)+c`

Step-by-step explanation

`\int \sin ^3(2x)\cos (2x)dx`

Step 1

substitute

Let

`\begin{gathered} u=\sin (2x) \\ \text{hence} \\ du=\cos (2x)\cdot2\cdot dx \\ du=2\cos (2x)dx \\ \text{divide both sides by 2} \\ (du)/(2)=(2\cos(2x))/(2)dx \\ (du)/(2)=\cos (2x)dx \end{gathered}`

Step 2

now, replace using the new variable(u)

`\begin{gathered} \int \sin ^3(2x)\cos (2x)dx \\ \int u^3(du)/(2) \end{gathered}`

take out the 1/2 from the integral

`\begin{gathered} \int u^3(du)/(2) \\ \int u^3(du)/(2)=(1)/(2)\int u^3du \\ (1)/(2)\int u^3du \end{gathered}`

finally, solve the simple integral

and rewrite using the original variable( x)

so

`\begin{gathered} (1)/(2)\int u^3du \\ (1)/(2)((u^4)/(4))+c \end{gathered}`

rewrite using x

`\begin{gathered} (1)/(2)((u^4)/(4))+c \\ (1)/(8)(u^4)+c \\ (1)/(8)(\sin ^4(2x))+c \\ (1)/(8)\sin ^4(2x)+c \end{gathered}`

therefore, the answer is

`(1)/(8)\sin ^4(2x)+c`

I hope this helps you