Integrate the following $int sqrt(x^2+81) dx$ ?

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Answer 1 · 2017-02-07T14:39:45

$(xsqrt(x^2 + 81))/2 + 81/2ln|(x + sqrt(x^2 + 81))/9| + C$

Use trig substitution. Let $x= 9tantheta$ . Then $dx = 9sec^2theta d theta$ .

$=>intsqrt((9tantheta)^2 + 81) * 9sec^2theta d theta$

$=>intsqrt(81tan^2theta + 81) * 9sec^2theta d theta$

$=>intsqrt(81(tan^2theta +1)) * 9sec^2theta d theta$

$=>intsqrt(81sec^2theta) * 9sec^2theta d theta$

$=>int 9sectheta * 9sec^2theta d theta$

$=>int 81sec^3theta d theta$

$=>81int sec^3theta d theta$

This is a relatively known integral. The proof can be found here.

$=>81/2secthetatantheta+ 81/2ln|sectheta+ tantheta| + C$

We know from our initial substitution that $tantheta = x/9$ . This means the hypotenuse of the triangle would have a hypotenuse of $sqrt(x^2 + 81)$ .

$=>81/2(sqrt(x^2 + 81)/9)(x/9) + 81/2ln|sqrt(x^2 + 81)/9 + x/9| + C$

$=>(xsqrt(x^2 + 81))/2 + 81/2ln|(x + sqrt(x^2 + 81))/9| + C$

Hopefully this helps!

Integrate the following int sqrt(x^2+81) dxint sqrt(x^2+81) dx?