Question

solve the integration of root (tanx)

Answer 1

$\frac{1}{\sqrt{2}} \arctan\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right) + \frac{1}{2\sqrt{2}} \ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right| + C$

Answer 2

To solve the integral $\int \sqrt{\tan x} \, dx$, we use a specific substitution and algebraic manipulation.

1. Substitution: Let $I = \int \sqrt{\tan x} \, dx$. Let $t^2 = \tan x$. Differentiating both sides with respect to $x$: $2t \frac{dt}{dx} = \sec^2 x$. We know that $\sec^2 x = 1 + \tan^2 x$. Substituting $t^2 = \tan x$, we get $\sec^2 x = 1 + (t^2)^2 = 1 + t^4$. So, $2t \frac{dt}{dx} = 1 + t^4$, which implies $dx = \frac{2t}{1+t^4} \, dt$.

Substitute $t^2 = \tan x$ and $dx = \frac{2t}{1+t^4} \, dt$ into the integral: $I = \int t \cdot \frac{2t}{1+t^4} \, dt = \int \frac{2t^2}{1+t^4} \, dt$.

2. Decompose the Integrand: We can split the integrand $\frac{2t^2}{t^4+1}$ into two parts: $\frac{2t^2}{t^4+1} = \frac{t^2+1 + t^2-1}{t^4+1} = \frac{t^2+1}{t^4+1} + \frac{t^2-1}{t^4+1}$. So, the integral becomes: $I = \int \frac{t^2+1}{t^4+1} \, dt + \int \frac{t^2-1}{t^4+1} \, dt$.

3. Evaluate the First Integral: Let $I_1 = \int \frac{t^2+1}{t^4+1} \, dt$. Divide the numerator and denominator by $t^2$: $I_1 = \int \frac{1 + 1/t^2}{t^2 + 1/t^2} \, dt$. Let $u = t - 1/t$. Then $du = (1 + 1/t^2) \, dt$. Also, $u^2 = (t - 1/t)^2 = t^2 + 1/t^2 - 2$, so $t^2 + 1/t^2 = u^2 + 2$. Substitute these into $I_1$: $I_1 = \int \frac{du}{u^2+2} = \int \frac{du}{u^2+(\sqrt{2})^2}$. This is a standard integral of the form $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \arctan\left(\frac{x}{a}\right)$. $I_1 = \frac{1}{\sqrt{2}} \arctan\left(\frac{u}{\sqrt{2}}\right) + C_1$. Substitute back $u = t - 1/t = \frac{t^2-1}{t}$: $I_1 = \frac{1}{\sqrt{2}} \arctan\left(\frac{t^2-1}{\sqrt{2}t}\right) + C_1$.

4. Evaluate the Second Integral: Let $I_2 = \int \frac{t^2-1}{t^4+1} \, dt$. Divide the numerator and denominator by $t^2$: $I_2 = \int \frac{1 - 1/t^2}{t^2 + 1/t^2} \, dt$. Let $v = t + 1/t$. Then $dv = (1 - 1/t^2) \, dt$. Also, $v^2 = (t + 1/t)^2 = t^2 + 1/t^2 + 2$, so $t^2 + 1/t^2 = v^2 - 2$. Substitute these into $I_2$: $I_2 = \int \frac{dv}{v^2-2} = \int \frac{dv}{v^2-(\sqrt{2})^2}$. This is a standard integral of the form $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right|$. $I_2 = \frac{1}{2\sqrt{2}} \ln\left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right| + C_2$. Substitute back $v = t + 1/t = \frac{t^2+1}{t}$: $I_2 = \frac{1}{2\sqrt{2}} \ln\left|\frac{\frac{t^2+1}{t}-\sqrt{2}}{\frac{t^2+1}{t}+\sqrt{2}}\right| + C_2 = \frac{1}{2\sqrt{2}} \ln\left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right| + C_2$.

5. Combine and Substitute Back: Combine $I_1$ and $I_2$: $I = I_1 + I_2 = \frac{1}{\sqrt{2}} \arctan\left(\frac{t^2-1}{\sqrt{2}t}\right) + \frac{1}{2\sqrt{2}} \ln\left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right| + C$. Finally, substitute back $t^2 = \tan x$ and $t = \sqrt{\tan x}$: $I = \frac{1}{\sqrt{2}} \arctan\left(\frac{\tan x-1}{\sqrt{2}\sqrt{\tan x}}\right) + \frac{1}{2\sqrt{2}} \ln\left|\frac{\tan x-\sqrt{2}\sqrt{\tan x}+1}{\tan x+\sqrt{2}\sqrt{\tan x}+1}\right| + C$.

The final answer is $\boxed{\frac{1}{\sqrt{2}} \arctan\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right) + \frac{1}{2\sqrt{2}} \ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right| + C}$.

Explanation of the solution:

1. Substitute $t^2 = \tan x$ to transform the integral into a rational function of $t$: $\int \frac{2t^2}{1+t^4} \, dt$.
2. Decompose the integrand $\frac{2t^2}{1+t^4}$ into $\frac{t^2+1}{t^4+1} + \frac{t^2-1}{t^4+1}$.
3. For $\int \frac{t^2+1}{t^4+1} \, dt$, divide numerator and denominator by $t^2$ and substitute $u=t-1/t$. This leads to an $\arctan$ function.
4. For $\int \frac{t^2-1}{t^4+1} \, dt$, divide numerator and denominator by $t^2$ and substitute $v=t+1/t$. This leads to a $\ln$ function.
5. Combine the results and substitute back $t = \sqrt{\tan x}$.

Answer: The integral of $\sqrt{\tan x}$ is $\frac{1}{\sqrt{2}} \arctan\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right) + \frac{1}{2\sqrt{2}} \ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right| + C$.