Hi can someone help integrate this function using trig sub for somereason i am off Plz show all steps
1/(1+9k^2)
This was answered but i am confused y u pull out a 1/3
1/(1+9k^2)
This was answered but i am confused y u pull out a 1/3
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(1 / { 9 [ (1/9) + k² ] }
I = ∫ (1 / { 9 [ (1/3)² + k² ] } dk
I = (1/9) 3 tan^(-1) [ 3k ] + C
I = (1/3) tan^(-1) [ 3k ] + C
I = ∫ (1 / { 9 [ (1/3)² + k² ] } dk
I = (1/9) 3 tan^(-1) [ 3k ] + C
I = (1/3) tan^(-1) [ 3k ] + C