How do you integrate #int 1/sqrt(-e^(2x) +49)dx# using trigonometric substitution?

Answer 1

#color(blue)(int 1/sqrt(49-e^(2x))dx=1/7*ln(7-sqrt(49-e^(2x)))-1/7x+C)#

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.

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Answer 2

To integrate ( \int \frac{1}{\sqrt{-e^{2x} + 49}} , dx ) using trigonometric substitution, let ( x = \frac{1}{2}\ln{\left(\frac{7}{\sin{\theta}}\right)} ). Then, ( dx = -\frac{1}{2}\csc{\theta}\cot{\theta} , d\theta ). Substitute these into the integral and simplify. This results in ( \int \frac{-1}{7\sqrt{1-\sin^2{\theta}}} \cdot \frac{1}{2}\csc{\theta}\cot{\theta} , d\theta ). This simplifies to ( \frac{-1}{14} \int \cot^2{\theta} , d\theta ). Rewrite ( \cot^2{\theta} ) as ( \csc^2{\theta} - 1 ) and integrate term by term. This yields ( \frac{-1}{14} \left( -\cot{\theta} - \theta \right) + C ). Substitute ( x ) back in terms of ( \theta ) to get the final answer: ( \frac{1}{14} \ln{\left( \frac{7}{\sin{\theta}} \right)} - \frac{x}{14} + C ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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