Unfortunately I am not able to solve the following integral:
Can somebody help me?
you can also use trigonometric substitutions.
One way is to draw a right-angled triangle with
The hypotenuse of this triangle is 2, since
Then, with the acute angle opposite the side of length 8x, we get
Shouldn't the answer be: 1/8 arcsin(4x)+c instead of arcsin(4x)+c?
@ archie meade
Thanks for your explanation about the application of trigonometric substitutions to this exercise. I didn't know I could use this for this exercise. Trigonometric substitutions are quite new to me.
yes, the trig substitutions can be tried later in your development.
It's best for you to master the inverse sine solution
recommended by the other contributors earlier.
dx needs to be changed to dy