The given #int 1/sqrt(49-e^(2x))dx#
Let #e^x=7*sin theta#
and #e^(2x)=49*sin^2 theta#
and #e^x dx=7*cos theta*d theta#
and #dx=(7*cos theta*d theta)/e^x=(7*cos theta*d theta)/(7*sin theta)=(cos theta*d theta)/(sin theta)#
Let us do the integration
#int 1/sqrt(49-e^(2x))dx=int 1/sqrt(49-49*sin^2 theta)*(cos theta*d theta)/(sin theta)#
#int 1/sqrt(49-e^(2x))dx=int 1/(sqrt(49)sqrt(1-sin^2 theta))*(cos theta*d theta)/(sin theta)#
#int 1/sqrt(49-e^(2x))dx=int 1/(7*sqrt(cos^2 theta))*(cos theta*d theta)/(sin theta)#
#int 1/sqrt(49-e^(2x))dx=int 1/(7*cos theta)*(cos theta*d theta)/(sin theta)#
#int 1/sqrt(49-e^(2x))dx=int 1/(7*cancelcos theta)*(cancelcos theta*d theta)/(sin theta)#
#int 1/sqrt(49-e^(2x))dx=1/7*int (d theta)/(sin theta)=1/7*int csc theta*d theta#
#int 1/sqrt(49-e^(2x))dx=1/7*ln(csc theta-cot theta)+C#
Now, we have to return it to the original variable x.
From our trigonometric substitution
#e^x=7*sin theta#
#sin theta=e^x/7# and our Right Triangle has opposite side#=e^x# and hypotenuse#=7# and adjacent side#=sqrt(49-e^(2x))#
Therefore,
#csc theta=7/e^x# and #cot theta=(sqrt(49-e^(2x)))/e^x#
And
#int 1/sqrt(49-e^(2x))dx=1/7*ln(csc theta-cot theta)+C#
#int 1/sqrt(49-e^(2x))dx=1/7*ln(7/e^x-(sqrt(49-e^(2x)))/e^x)+C#
#int 1/sqrt(49-e^(2x))dx=1/7*ln((7-sqrt(49-e^(2x)))/e^x)+C#
#int 1/sqrt(49-e^(2x))dx=1/7*[ln(7-sqrt(49-e^(2x)))-ln e^x]+C#
#int 1/sqrt(49-e^(2x))dx=1/7*ln(7-sqrt(49-e^(2x)))-1/7x+C#
God bless....I hope the explanation is useful.
