How does one plug radicals with non-perfect squares and variables into the Pythagorean theorem formula? I am working on the following integral 
$$\int\left( 7x^2 - 3 \right)^{\frac 5 2} \, dx$$
I want to use the $\sqrt{u^2 - a^2}$ 
$u = a\sec\theta$
I know in order to get it into the form that will allow me to do this I will have to do the following.
$$\int \sqrt {\left( 7{ x }^{ 2 }-3 \right)^5 } \, dx$$
At this point in solving this kind of problems I would draw a right triangle and place the values for a b and c in the appropriate places. This allows me to see that the substitution I am choosing is going to work and this lets me have something to refer to when putting the solved integral back into terms of x from theta. This usually works great when I am working with values such as 
$x^2-16$ 
or
$x^2-25$
These values plug into the the Pythagorean theorem formula just fine. However, the above produces radicals I am not sure how to deal with.
Triangle diagram
I am assuming the above drawing is correct.
I know regardless that I will end up substituting x with
$$x =\sqrt{\frac{3}{7}}\sec\theta$$
I know this is simple but I am not sure if I am understanding this correctly. Have I drawn this diagram correctly outside of the placement of $\sqrt{7x^2-3} $? Also, how does the math work exactly when plugging these values into the Pythagorean theorem?  I am not understanding how when I square $\sqrt{7x}$ I end up with $7x^2$. Shouldn't I end up with just $7x$? 
 A: "Have I drawn this diagram correctly outside of the placement of $\sqrt{7x^2-3} $?" 
The answer to this is no. The diagram that I drew was incorrect because, if one looks at the hypotenuses in the diagram, I am including the x variable WITHIN the square root as such $\sqrt{7x}$. This is not how a "non-perfect squared number with variable such as $(7x^2)$" should be transposed to the right triangle diagram from the square root below
$$\int \sqrt {\left( 7{ x }^{ 2 }-3 \right)^5 } \, dx$$
This is because, ONLY the coefficient and NOT the variable should be within the square root. In others words it should be written as the variable being multiplied by the square root of the coefficient as such $x\sqrt{7}$.   
.
.
"Also, how does the math work exactly when plugging these values into the Pythagorean theorem? I am not understanding how when I square $\sqrt{7x}$ I end up with $7x^2$. Shouldn't I end up with just $7x$?"
The reason of the confusion here was because, when I was plugging the values of the diagram into the Pythagorean theorem I was running into $\sqrt{7x} * \sqrt{7x} = 7x$. This was because, just as in the first question answered, I was not transposing $7x^2$ to the triangle correctly like so $x\sqrt{7}$. This would give me $x\sqrt{7} * x\sqrt{7} = 7x^2$ which would be the correct logic to allow me to proceed with the problem and DERIVE the correct trigonometric substitution which is that below.
$$x =\sqrt{\frac{3}{7}}\sec\theta$$
A: Rather than focusing on the triangle, I usually first compare the integrand one of three Pythagorean identities:
$$\tan^2(\theta) + 1 = \sec^2(\theta)$$
$$1 - \sin^2(\theta) = \cos^2(\theta)$$
$$\sec^2(\theta) - 1 = \tan^2(\theta)$$
If the integrand is of the form $9x^2 + 1$ then I compare it with the first, and choose the substition $3x= \tan(\theta)$. For your integrand, $7x^2-3$, $\sec(\theta)$ seems to be the way to go. In particular, writing it as $$3 \left( \frac{7}{3} x^2 - 1 \right)$$ we see that the substitution $\sqrt{\frac{7}{3}} x = \sec(\theta)$ would be profitable. Moving forward with this choice we find that $$\sqrt{\frac73} dx = \sec(\theta)\tan(\theta) d\theta.$$
Proceeding with the substitutions we find $$\int (7x^2 - 3)^{\frac52} dx = 3^{\frac52} \sqrt{\frac{3}{7}} \int \left( \sec^2(\theta) - 1\right)^{\frac52} \sec(\theta)\tan(\theta) d \theta$$
$$=3^{\frac52} \sqrt{\frac{3}{7}} \int \tan^5(\theta) \sec(\theta)\tan(\theta) d \theta = \frac{3^{6/2}}{7^{1/2}} \int \tan^6(\theta) \sec(\theta) d \theta.$$
We are left now with the integral $$\int \tan^6(\theta) \sec(\theta) d\theta = \int ( \sec^2(\theta) - 1)^3 \sec(\theta) d\theta$$
$$=\int (\sec^6(\theta) - 3\sec^4(\theta) + 3\sec^2(\theta) - 1)\sec(\theta) d\theta$$
$$=\int \sec^7(\theta) - 3\sec^5(\theta) + 3\sec^3(\theta) - \sec(\theta) d\theta.$$
The integrals for $\sec(\theta)$ and $\sec^3(\theta)$ can be found in many standard integration tables.
You can find $\int \sec^7(\theta) d\theta$ depends on $\int \sec^5(\theta)$ after using integration by parts $$\int \sec^7(\theta) d\theta = \int \sec^2(\theta) \sec^5(\theta) d\theta = \sec^5(\theta) \cdot \tan(\theta) - \int \tan(\theta) (5 \sec^4(\theta) \sec(\theta)\tan(\theta)) d\theta$$
$$= \sec^5(\theta)\tan(\theta) - 5\int \tan^2(\theta) \sec^5(\theta) d\theta$$
$$= \sec^5(\theta)\tan(\theta) - 5\int (\sec^2(\theta)-1) \sec^5(\theta) d\theta$$
$$= \sec^5(\theta)\tan(\theta) - 5\int \sec^7(\theta)d\theta + 5\int \sec^5(\theta) d\theta.$$
Solving for the integral,
$$\int \sec^7(\theta) d\theta = \frac16 \sec^5(\theta) \tan(\theta) + \frac56 \int \sec^5(\theta) d\theta.$$
In a similar fashion you can show that $\int \sec^5(\theta) d\theta$ depends on $\int \sec^3(\theta) d\theta$.
Finally, when it comes time to substitute back in the $x$ variables, recall that we used $$\sqrt{\frac{7}{3}} x = \sec(\theta).$$ The secant of an angle in a right triangle is given by the ratio $\frac{hyp}{adj}$. Taking $\sqrt{7/3} x$ as our hypotenuse and $1$ as the adjacent side we find that the opposite side of the angle would be $$\sqrt{ \frac73 x^2 - 1}$$ which is very similar to what was in the integrand to begin with.
A: Hint:
use as substitution hyperbolic  cosine  $$ x=\sqrt { \frac { 3 }{ 7 }  } \cosh { t } ,dx=\sqrt { \frac { 3 }{ 7 }  } \sinh { t } dt$$
$$\\ \\ \frac { 27 }{ \sqrt { 7 }  } \int { \sqrt { { \left( \cosh ^{ 2 }{ t } -1 \right)  }^{ 5 } }  } \sinh { t } dt=\frac { 27 }{ \sqrt { 7 }  } \int { \sinh ^{ 6 }{ t }  } dt$$
