How do you find the indefinite integral of #sqrt(25 + x^2)#?

Question

Charles Arocha · Accepted Answer

#1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2)))+C#

Use trigonometric substitution.

Draw a triangle with an angle #theta#.

Label the opposite side as #x# and adjacent side as #5#.

Now examine the triangle and notice that

#tantheta=x/5#

#rArrx=5tantheta#

#rArrdx=5sec^2theta# #d##theta#

By the Pythagorean Theorem, the hypotenuse of the triangle is:

#sqrt(25+x^2)#

So we can write

#sectheta=sqrt(25+x^2)/5#

#rArrsqrt(25+x^2)=5sectheta#

Let's now rewrite the integral in terms of #theta#

#intsqrt(25+x^2)dxrArrint(5sectheta)(5sec^2theta# #d##theta)#

#rArr25int# #sec^3theta# #d##theta#

Integrating #sec^3theta# can be tricky. There is a shortcut formula, but I'll do it the long way for teaching purposes.

Separate #(sec^3theta# #d##theta)# into #(sectheta*sec^2theta# #d##theta)# and use integration by parts.

Let:

#u=sectheta# #du=secthetatantheta# #d##theta#

#dv=sec^2theta# #d##theta# #v=tantheta#

Then:

#rArrint# #sec^3theta# #d##theta=secthetatantheta-int# #tan^2thetasectheta# #d##theta#

#rArrint# #sec^3theta# #d##theta=secthetatantheta-int# #(sec^2theta-1)sectheta# #d##theta#

#rArrint# #sec^3theta# #d##theta=secthetatantheta-int# #(sec^3theta-sectheta)##d##theta#

#rArr2int# #sec^3theta# #d##theta=secthetatantheta+int# #sectheta# #d##theta#

#rArr2int# #sec^3theta# #d##theta=secthetatantheta+lnabs(sectheta+tantheta)#

#rArrint# #sec^3theta# #d##theta=1/2(secthetatantheta+lnabs(sectheta+tantheta))+C#

And let's not forget that our integral was multiplied by 25!

#rArr25int# #sec^3theta# #d##theta=25/2(secthetatantheta+lnabs(sectheta+tantheta))+C#

Now put everything back in terms of #x#

#rArr25/2(sqrt(25+x^2)/5x/5+lnabs(sqrt(25+x^2)/5+x/5))+C#

#rArr1/2(xsqrt(25+x^2)+25lnabs((sqrt(25+x^2)+x)/5))+C#

#rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2))-25ln5)+C#

Then we can absorb the constant #-(25ln5)/2# into #C# to get our final answer:

#rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2)))+C#

Christopher Agresta · Answer

undefined To find the indefinite integral of sqrt(25 + x^2), use the trigonometric substitution method. Let $ x = 5 \tan(\theta) $, then $ dx = 5 \sec^2(\theta) d\theta $. Substitute these into the integral, and simplify using trigonometric identities. The integral becomes $ \int 25 \sec^3(\theta) d\theta $. Then use integration techniques for trigonometric functions to solve this integral. The result will involve trigonometric functions and constants.

Charles Anctil · Answer

undefined To find the indefinite integral of $ \sqrt{25 + x^2} $, we can use a trigonometric substitution.

Let $ x = 5 \tan(\theta) $, then $ dx = 5 \sec^2(\theta) d\theta $.

Substituting $ x $ and $ dx $ into the integral, we get:

$ \int \sqrt{25 + x^2} \, dx = \int \sqrt{25 + (5\tan(\theta))^2} \cdot 5\sec^2(\theta) \, d\theta $

$ = \int \sqrt{25 + 25\tan^2(\theta)} \cdot 5\sec^2(\theta) \, d\theta $

$ = \int \sqrt{25(1 + \tan^2(\theta))} \cdot 5\sec^2(\theta) \, d\theta $

$ = \int \sqrt{25\sec^2(\theta)} \cdot 5\sec^2(\theta) \, d\theta $

$ = \int 5\sec^3(\theta) \, d\theta $

Now, we can integrate $ \int 5\sec^3(\theta) \, d\theta $ using a standard integral formula.

The result is:

$ = \frac{5}{2} (\sec(\theta) \tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|) + C $

Finally, we substitute back $ \theta $ using the inverse trigonometric function, $ \theta = \arctan\left(\frac{x}{5}\right) $.

So, the indefinite integral of $ \sqrt{25 + x^2} $ is:

$ \frac{5}{2} (\sec(\arctan(x/5)) \tan(\arctan(x/5)) + \ln|\sec(\arctan(x/5)) + \tan(\arctan(x/5))|) + C $

This expression can be simplified further, but the above form represents the indefinite integral of $ \sqrt{25 + x^2} $.