# Math Help - trigonometry substitution

1. ## trigonometry substitution

$\int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx$

I draw the 146 outside the integral and set the substitution as;

6x = 11 sec t
x = 11/6 sec t
dx = 11/6 sec t tan t

I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

$146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}$

Is this correct?

2. Originally Posted by Archduke01
$\int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx$

I draw the 146 outside the integral and set the substitution as;

6x = 11 sec t
x = 11/6 sec t
dx = 11/6 sec t tan t

I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

$146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}$

Is this correct?
$x^2=\frac{121}{36}sec^2t$

You also forgot the -121 under the square root

3. Originally Posted by Archduke01
$\int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx$

I draw the 146 outside the integral and set the substitution as;

6x = 11 sec t
x = 11/6 sec t
dx = 11/6 sec t tan t

I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

$146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}$

Is this correct?
$6x = 11\sec{t}$

$36x^2 = 121\sec^2{t}$

$x^2 = \frac{121}{36}\sec^2{t}$

$dx = \frac{11}{6}\sec{t}\tan{t} \, dt
$

$146 \int \frac{\frac{11}{6}\sec{t}\tan{t}}{\frac{121}{36}\s ec^2{t}\sqrt{121\sec^2{t} - 121}} \, dt$

$146 \int \frac{\frac{1}{6}\sec{t}\tan{t}}{\frac{121}{36}\se c^2{t}\sqrt{\sec^2{t} - 1}} \, dt$

$146 \cdot \frac{1}{6} \cdot \frac{36}{121} \int \frac{\sec{t}\tan{t}}{\sec^2{t}\sqrt{\sec^2{t} - 1}} \, dt$

$\frac{876}{121} \int \cos{t} \, dt
$

finish?

4. Originally Posted by Archduke01
$\int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx$

I draw the 146 outside the integral and set the substitution as;

6x = 11 sec t
x = 11/6 sec t
dx = 11/6 sec t tan t

I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

$146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}$

Is this correct?
Hyperbolic substitution works well in this case.

$\int{\frac{146}{x^2\sqrt{36x^2 - 121}}\,dx} = \int{\frac{146}{6x^2\sqrt{x^2 - \frac{121}{36}}}\,dx}$

$= \int{\frac{73}{3x^2\sqrt{x^2 - \frac{121}{36}}}\,dx}$

Now make the substitution $x = \frac{11}{6}\cosh{t}$ so that $dx = \frac{11}{6}\sinh{t}\,dt$.

The integral becomes:

$\int{\frac{73}{3\left(\frac{11}{6}\cosh{t}\right)^ 2\sqrt{\left(\frac{11}{6}\cosh{t}\right)^2 - \frac{121}{36}}}\,\frac{11}{6}\sinh{t}\,dt}$

$= \int{\frac{803\sinh{t}}{18\left(\frac{121}{36}\cos h^2{t}\right)\sqrt{\frac{121}{36}\cosh^2{t} - \frac{121}{36}}}\,dt}$

$= \int{\frac{803\sinh{t}}{\frac{121}{2}\cosh^2{t}\cd ot \frac{11}{6}\sqrt{\cosh^2{t} - 1}}\,dt}$

$= \int{\frac{803\sinh{t}}{\frac{1331}{12}\cosh^2{t}\ sqrt{\sinh^2{t}}}\,dt}$

$= \frac{876}{121}\int{\frac{\sinh{t}}{\cosh^2{t}\sin h{t}}\,dt}$

$= \frac{876}{121}\int{\frac{1}{\cosh^2{t}}\,dt}$

$= \frac{876}{121}\int{\textrm{sech}^2{t}\,dt}$

$= \frac{876}{121}\tanh{t} + C$

Now remembering that we made the substitution $x = \frac{11}{6}\cosh{t}$

$\cosh{t} = \frac{6x}{11}$.

Also, since $\tanh{t} = \frac{\sqrt{\cosh^2{t} - 1}}{\cosh{t}}$

$\tanh{t} = \frac{\sqrt{\left(\frac{6x}{11}\right)^2 - 1}}{\frac{6x}{11}}$

$= \frac{11\sqrt{\frac{36x^2 - 121}{121}}}{6x}$

$= \frac{\sqrt{36x^2 - 121}}{6x}$

Therefore, our integral, which we got to

$\frac{876}{121}\tanh{t} + C$

becomes

$\frac{876}{121}\left(\frac{\sqrt{36x - 121)}}{6x}\right) + C$

$= \frac{146\sqrt{36x - 121}}{121x} + C$.