Question

Find
∫sec3xtanxdx

Answer 1

Answer on Question #45969 – Math – Integral Calculus

1. Find $\int \sec 3x \tan x \, dx$.

Solution.

First of all we will rewrite our integral:

$$\int \sec 3x \tan x \, dx = \int \frac{\tan x}{\cos 3x} \, dx = \int \frac{\sin x}{\cos 3x \cos x} \, dx.$$

Now we can use formula: $\cos 3x = 4\cos^3 x - 3\cos x$. Then we will use substitution:

$$t = \cos^2 x,$$ $$x = \arccos \sqrt{t},$$ $$dx = -\frac{dt}{2\sqrt{t}\sqrt{1-t}}.$$

After this we will have:

$$\begin{aligned} \int \frac{\sin x}{\cos 3x \cos x} \, dx &= \int \frac{\sin x}{(4\cos^3 x - 3\cos x)\cos x} \, dx = \int \frac{\sqrt{1 - \cos^2 x}}{4\cos^4 x - 3\cos^2 x} \, dx = \\ &= -\frac{1}{2} \int \frac{\sqrt{1-t}}{\sqrt{t}(4t^2 - 3t)\sqrt{1-t}} \, dt = -\frac{1}{2} \int \frac{1}{t^2(4t - 3)} \, dt = -\frac{1}{2} \int t^{-\frac{3}{2}}(4t - 3)^{-1} \, dt. \end{aligned}$$

We obtained the Chebyshev integral:

$$\int x^m (a + b x^n)^p \, dx.$$

When $p$ is integer, as in our case, we can do substitution $x = t^s$, where $s$ is least common denominator of $m$ and $n$.

In our case $m = -\frac{3}{2}$ and $n = 1$. Hence, least common denominator is 2. And now we can do substitution $t = y^2$ and $dt = 2ydy$:

$$\begin{aligned} & -\frac{1}{2} \int t^{-\frac{3}{2}}(4t - 3)^{-1} \, dt = -\frac{1}{2} \int y^{-3}(4y^2 - 3)^{-1} \, 2ydy = -\int \frac{1}{y^2(4y^2 - 3)} \, dy = \\ & = \frac{1}{3} \int \frac{1}{y^2\left(1 - \frac{4}{3}y^2\right)} \, dx. \end{aligned}$$

We can rewrite our fraction:

$$\frac{1}{y^2\left(1 - \frac{4}{3}y^2\right)} = \frac{1}{y^2} + \frac{\frac{4}{3}}{1 - \frac{4}{3}y^2}.$$

Hence, we will have two integrals:

$$\frac{1}{3} \int \frac{1}{y^2} \, dy + \frac{4}{9} \int \frac{1}{1 - \frac{4}{3}y^2} \, dy = -\frac{1}{3y} + \frac{4}{9} \sqrt{\frac{3}{4}} \operatorname{arctanh} \sqrt{\frac{4}{3}} y.$$

Now we must return to $x$:

$$\begin{array}{l} - \frac {1}{3 y} + \frac {4}{9} \sqrt {\frac {3}{4}} \operatorname {arctanh} \left(\frac {4}{3} y\right) = - \frac {1}{3 \sqrt {t}} + \frac {2}{3 \sqrt {3}} \operatorname {arctanh} \left(\frac {2}{\sqrt {3}} \sqrt {t}\right) = - \frac {1}{3 \cos x} + \\ + \frac {2}{3 \sqrt {3}} \operatorname {arctanh} \left(\frac {2}{\sqrt {3}} \cos x\right). \end{array}$$

Answer:

$$\text{result is} - \frac {1}{3 \cos x} + \frac {2}{3 \sqrt {3}} \operatorname {arctanh} \left(\frac {2}{\sqrt {3}} \cos x\right).$$

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