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int[sqrt(tan x)]dx

[tanx]dx \int[\sqrt{\tan x}] d x

Full solution

Q. [tanx]dx \int[\sqrt{\tan x}] d x
  1. Simplify Integral Expression: Step 11: Simplify the integral expression.\newlineWe start by recognizing that the integral of the square root of the tangent function is not straightforward. We need to simplify or find a substitution that makes it easier to integrate.\newlineCalculation: tanxdx\int \sqrt{\tan x}\,dx
  2. Choose Substitution: Step 22: Choose a substitution.\newlineLet u=tan(x)u = \tan(x), then du=sec2(x)dxdu = \sec^2(x) dx. This substitution will transform the integral into a form involving uu, which might be easier to integrate.\newlineCalculation: dx=dusec2(x)dx = \frac{du}{\sec^2(x)}
  3. Rewrite Using Substitution: Step 33: Rewrite the integral using the substitution.\newlineSubstituting the values from Step 22, we get:\newline[tanx]dx=[u(1/sec2(x))]du\int[\sqrt{\tan x}]\,dx = \int[\sqrt{u} \cdot (1/\sec^2(x))] \,du\newlineSince sec2(x)=1+tan2(x)=1+u2\sec^2(x) = 1 + \tan^2(x) = 1 + u^2, we have:\newline[u(1/(1+u2))]du\int[\sqrt{u} \cdot (1/(1 + u^2))] \,du
  4. Attempt Integration: Step 44: Attempt to integrate the transformed expression.\newlineThis integral looks complex and might not have a straightforward antiderivative in terms of elementary functions. We attempt integration:\newlineu1+u2du\int \frac{\sqrt{u}}{1 + u^2} \, du