Integration involving trigonometric substitutions

[paulmdrdo1]

Please tell me if i worked out the problem correctly.

1. ∫(dx/x²-x+2)

completing the square of the denominator i have,

∫[dx/(x-1)²+1²]

a=1, u=x-1; du=dx

∫(du/a²+u²)

1/1*tan^-1x-1/1 + c = tan^-1x-1 + C -->> final answer

2. ∫[dx/(15+2x-x²)^1/2]

∫[dx/(14-(x-1)²)^1/2]

∫{dx/([(14)^1/2]²-(x-1)²)}

a = (14)^1/2; u = x-1; du = dx

= 1/(14)^1/2*sin^-1x-1/(14)^1/2 + C --->>final answer

and P.S how to use that font that you are using?

 

[Opalg]

Re: Integration of inverse Trig functions

Please tell me if i worked out the problem correctly.

1. ∫(dx/x²-x+2)

completing the square of the denominator i have,

∫[dx/(x-1)²+1²]

a=1, u=x-1; du=dx

∫(du/a²+u²)

1/1*tan^-1x-1/1 + c = tan^-1x-1 + C -->> final answer

Method is right, algebra is wrong. In fact, $x^2-x+2 = \bigl(x-\frac12\bigr)^2 + \frac74$.

and P.S how to use that font that you are using?

To typeset those fonts you need to learn http://www.mathhelpboards.com/f26/math-help-boards-latex-guide-pdf-1142/.

 

[paulmdrdo1]

Re: Integration of inverse Trig functions

my new answer to 2 is this 2/(7)^1/2*tan^-12x-1/(7)^1/2

are my answers to 1 and 2 now correct?

 

[MarkFL]

Re: Integration of inverse Trig functions

In your first attempt at 2.) you completed the square incorrectly:

\(\displaystyle 15+2x-x^2=16-(x-1)^2=4^2-(x-1)^2\)

 

[paulmdrdo1]

Re: Integration of inverse Trig functions

1/4*sin^-1x-1/4 + C --->> is this the correct answer for 1?

2/(7)^1/2*tan^-12x-1/(7)^1/2 ---> answer for 2.

are they now correct? sorry for not typing my solution. I'm in a rush.

 

[Prove It]

Re: Integration of inverse Trig functions

Please tell me if i worked out the problem correctly.

1. ∫(dx/x²-x+2)

completing the square of the denominator i have,

∫[dx/(x-1)²+1²]

a=1, u=x-1; du=dx

∫(du/a²+u²)

1/1*tan^-1x-1/1 + c = tan^-1x-1 + C -->> final answer

2. ∫[dx/(15+2x-x²)^1/2]

∫[dx/(14-(x-1)²)^1/2]

∫{dx/([(14)^1/2]²-(x-1)²)}

a = (14)^1/2; u = x-1; du = dx

= 1/(14)^1/2*sin^-1x-1/(14)^1/2 + C --->>final answer

and P.S how to use that font that you are using?

Since the OP has gotten the correct answer to Q1 and is very close to getting the correct answer to Q2, I will post how I would approach these problems, which involves trigonometric or hyperbolic substitutions.

Q1.

\(\displaystyle \displaystyle \begin{align*} \int{ \frac{dx}{x^2 - x + 2}} &= \int{ \frac{dx}{ \left( x - \frac{1}{2} \right) ^2 + \frac{7}{4} } } \end{align*}\)

Now make the substitution \(\displaystyle \displaystyle \begin{align*} x - \frac{1}{2} = \frac{\sqrt{7}}{2}\tan{(\theta)} \implies dx = \frac{\sqrt{7}}{2}\sec^2{(\theta)}\,d\theta \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{ \frac{dx}{ \left( x - \frac{1}{2} \right) ^2 + \frac{7}{4} } } &= \int{ \frac{\frac{\sqrt{7}}{2}\sec^2{(\theta)}\,d\theta}{ \left[ \frac{\sqrt{7}}{2}\tan{(\theta)} \right] ^2 + \frac{7}{4} } } \\ &= \frac{\sqrt{7}}{2} \int{ \frac{\sec^2{(\theta)}\,d\theta}{\frac{7}{4}\tan^2{(\theta)} + \frac{7}{4}} } \\ &= \frac{\sqrt{7}}{2}\cdot \frac{4}{7} \int{\frac{\sec^2{(\theta)}\,d\theta}{\tan^2{( \theta )} + 1}} \\ &= \frac{2\sqrt{7}}{7} \int{ \frac{\sec^2{(\theta)}\,d\theta}{\sec^2{(\theta)}} } \\ &= \frac{2\sqrt{7}}{7}\int{1\,d\theta} \\ &= \frac{2\sqrt{7}}{7} \theta + C \\ &= \frac{2\sqrt{7}}{7}\arctan{\left[ \frac{2\sqrt{7}}{7}\left( x - \frac{1}{2} \right) \right] } + C \end{align*}\)

Q2.

\(\displaystyle \displaystyle \begin{align*} \int{\frac{dx}{\sqrt{15 + 2x - x^2}}} &= \int{\frac{dx}{\sqrt{-\left( x^2 - 2x - 15 \right) } } } \\ &= \int{ \frac{dx}{\sqrt{-\left[ x^2 - 2x + (-1) ^2 - (-1)^2 - 15 \right] }}} \\ &= \int{ \frac{dx}{\sqrt{- \left[ (x - 1)^2 - 16 \right] }}} \\ &= \int{ \frac{dx}{\sqrt{16 - (x - 1)^2}}} \end{align*}\)

Now make the substitution \(\displaystyle \displaystyle \begin{align*} x - 1 = 4\sin{(\theta)} \implies dx = 4\cos{(\theta)}\,d\theta \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{\frac{dx}{\sqrt{16 - (x-1)^2}}} &= \int{\frac{4\cos{(\theta)}\,d\theta}{\sqrt{16 - \left[ 4\sin{(\theta)} \right] ^2 }}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{16 - 16\sin^2{(\theta)}}}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{16 \left[ 1 - \sin^2{(\theta)} \right] }}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{16\cos^2{(\theta)}}}} \\ &= 4\int{\frac{\cos{(\theta)}\,d\theta}{4\cos{(\theta)}}} \\ &= \int{ 1\,d\theta} \\ &= \theta + C \\ &= \arcsin{\left[ \frac{1}{4} \left( x - 1 \right) \right] } + C \end{align*}\)

 

[paulmdrdo1]

Re: Integration of inverse Trig functions

how did you know this substitution?
\begin{align*} x - \frac{1}{2} = \frac{\sqrt{7}}{2}\tan{(\theta)} \implies dx = \frac{\sqrt{7}}{2}\sec^2{(\theta)}\,d\theta \end{align*}

why is \begin{align*} x - \frac{1}{2} equal to \frac{\sqrt{7}}{2}\tan{(\theta)}\end{align*} ?

please tell me. i want to know different techniques.

 

[MarkFL]

Re: Integration of inverse Trig functions

Consider the Pythagorean identities:

\(\displaystyle \sin^2(\theta)+\cos^2(\theta)=1\)

\(\displaystyle \tan^2(\theta)+1=\sec^2(\theta)\)

\(\displaystyle 1+\cot^2(\theta)=\csc^2(\theta)\)

You want to use a substitution that allows you to utilize one of these identities. In the problem to which you refer, the integrand is:

\(\displaystyle \frac{1}{\left(x-\frac{1}{2} \right)^2+\frac{7}{4}}\)

We should observe that either the second or third identity above will be suitable. Like Prove It, I would choose the third. And so we wish to have:

\(\displaystyle \left(x-\frac{1}{2} \right)^2=\frac{7}{4}\tan^2(\theta)\)

This way, the \(\displaystyle \frac{7}{4}\) may be factored out, leaving us with an expression to which we may apply the third Pythagorean identity above. So, taking the positive root, we find::

\(\displaystyle x-\frac{1}{2}=\frac{\sqrt{7}}{2}\tan(\theta)\)

and differentiating, we find:

\(\displaystyle dx=\frac{\sqrt{7}}{2}\sec^2(\theta)\,d\theta\)

and now our integral is:

\(\displaystyle \frac{2}{\sqrt{7}}\int\frac{\sec^2(\theta)}{\tan^2(\theta)+1}\,d\theta\)

Using the Pythagorean identity, we find we have:

\(\displaystyle \frac{2}{\sqrt{7}}\int\frac{\sec^2(\theta)}{\sec^2(\theta)}\,d\theta=\frac{2}{\sqrt{7}}\int\,d\theta=\frac{2}{\sqrt{7}}\theta+C\)

Now we need to back substitute for $\theta$, and so we take:

\(\displaystyle x-\frac{1}{2}=\frac{\sqrt{7}}{2}\tan(\theta)\)

and solve for $\theta$:

\(\displaystyle \frac{2x-1}{\sqrt{7}}=\tan(\theta)\)

\(\displaystyle \theta=\tan^{-1}\left(\frac{2x-1}{\sqrt{7}} \right)\)

And now we may conclude:

\(\displaystyle \int\frac{dx}{x^2-x+2}=\frac{2}{\sqrt{7}}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}} \right)+C\)

 

[paulmdrdo1]

uhmm. i still could not see the relation of this to that identity that you use.
can you show me explicitly how would you arrive to this
\begin{align*} x - \frac{1}{2} equal to \frac{\sqrt{7}}{2}\tan{(\theta)}\end{align*}

 

[Prove It]

First of all, please use an equals sign "=" instead of writing "equals to".

Second, you are allowed to substitute in ANY function that you like. We make this choice because we know that the function chosen simplifies with the Pythagorean Identity, and we also know that this simplification is very similar (in fact, identical) to the derivative of the substituted function, which means it will cancel, leaving a VERY easy function to integrate in terms of the new variable.