Solve Integral: $\int \sqrt{\tan(x)} \ dx$

[j_madison]

<br /> \int \sqrt{tan(x)} dx<br />

this problem has me stumped. any suggestions would be greatly appreciated. thanks

 

[devious_]

I remember doing this a while back. It's pretty long-winded. Anyway, here's a push in the right direction:

Use the substitution u=\sqrt{\tan(x)}, then:
u^2=\tan(x)[/itex]<br /> 2u\,du=\sec^2(x)\,dx<br /> 2u\,du=1+\tan^2(x)\,dx<br /> dx = \frac{2u}{1+u^4}\,du

 

[M98Ranger]

Hi there,

Solving integrals can be a bit challenging, but with some practice and understanding of the different techniques, it can become easier. In this case, we have an integral of the form $\int \sqrt{\tan(x)} \ dx$.

One approach to solve this integral is by using the substitution method. Let's assume $u = \tan(x)$, then $du = \sec^2(x) dx$. We can rewrite the integral as:

$$\int \sqrt{\tan(x)} \ dx = \int \sqrt{u} \ \frac{du}{\sec^2(x)} = \int u^{\frac{1}{2}} \cos^2(x) \ du$$

Now, we can use the trigonometric identity $\cos^2(x) = \frac{1}{2} (1 + \cos(2x))$ to rewrite the integral as:

$$\int u^{\frac{1}{2}} \cos^2(x) \ du = \frac{1}{2} \int u^{\frac{1}{2}} (1 + \cos(2x)) \ du$$

Using the power rule for integration, we get:

$$\frac{1}{2} \int u^{\frac{1}{2}} (1 + \cos(2x)) \ du = \frac{1}{2} \left(\frac{2}{3} u^{\frac{3}{2}} + \frac{1}{2} \sin(2x) \right) + C$$

Substituting back $u = \tan(x)$, we finally get:

$$\int \sqrt{\tan(x)} \ dx = \frac{1}{3} \tan^{\frac{3}{2}}(x) + \frac{1}{4} \sin(2x) + C$$

And that's it! We have successfully solved the integral. I hope this helps and good luck with your future math problems!