Question

integrate 1 / sqrt(25 - 16x^2) dx

My answer was arcsin(4x/5) + c

However, others give answer as 1/4 arcsin(4x/5) + c. I am not sure where the 1/4 comes from. I see its the sqrt of 16, but I'm not sure the rules that make it part of the answer. Am I just missing something easy?

Answer 1

check your answer by differentiating and you will see what the constant should be

Answer 2

*coefficient

Answer 3

so the only way to know that is to take it back the other way once I have applied the formula?

Answer 4

no i am just saying that it will make sense when you differentiate. you are looking for a "rule" but you will see for yourself what you need to do to ensure that the function you get had the derivative you want

Answer 5

ok, i see I need to cancel out the 4. Does this simply apply to all coefficients of the square term?

Answer 6

\[\int\limits_{?}^{?}1/\sqrt{25-16x ^{2}}dx\]
\[\int\limits_{?}^{?}1/(5\sqrt{1-16x ^{2}/25})dx\]
\[\int\limits_{?}^{?}1/(5\sqrt{1-(4x/5 )^{2}})dx\]
\[1/4\int\limits_{?}^{?}1/\sqrt{1-(4x/5 )^{2}}4dx/5\]
change variable
\[1/4\int\limits_{?}^{?}1/\sqrt{1-(y )^{2}}dy\]
1/4 arcsin(4x/5)

Answer 7

thank you.