Solving ∫ x 1 x 2 d x E − U 0 tan ⁡ ( a x ) 2

Question

Solving        ∫                  x                  1                                    x                  2                                d      x              E      −              U                  0                    tan      ⁡      (      a      x              )                  2

Marisol Morton · Accepted Answer

Well, we are trying to find the following integral:

\begin{matrix} (1) & I_{n} (α, β) = \int \frac{1}{\sqrt{α + β \tan^{2} (n x)}} d x \end{matrix}

Let's substitute

u = n x

, this leads to:

\begin{matrix} (2) & I_{n} (α, β) = \frac{1}{n} \int \frac{1}{\sqrt{α + β \tan^{2} (u)}} du \end{matrix}

Let's substitute

s = \sin (u)

, this leads to:

\begin{matrix} (3) & I_{n} (α, β) = \frac{1}{n} \int \frac{1}{\sqrt{α + s^{2} (β - α)}} ds \end{matrix}

Let's substitute

w = s \cdot \sqrt{\frac{β}{α} - 1}

, this leads to:

\begin{matrix} (4) & I_{n} (α, β) = \frac{1}{n \sqrt{β - α}} \int \frac{1}{\sqrt{w^{2} + 1}} dw \end{matrix}

This is a very standard integral, which gives:

\begin{matrix} (5) & I_{n} (α, β) = \frac{1}{n \sqrt{β - α}} \cdot \ln | w + \sqrt{w^{2} + 1} | + C \end{matrix}

So, we end up with:

\begin{matrix} (6) & I_{n} (α, β) = \frac{1}{n \sqrt{β - α}} \cdot \ln | \sin (n x) \cdot \sqrt{\frac{β}{α} - 1} + \sqrt{\sin^{2} (n x) \cdot | \frac{β}{α} - 1 | + 1} | + C \end{matrix}

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