Trigonometric integral with cosinus

Question

I cannot solve the equation below: I know what I typed is wrong byt I can't understand where it went sour.

$$\int \frac{1}{\cos(5x)} dx$$ $$\left[ \begin{align*} 5x=s\\ x=\frac{s}{5}; \frac{dx}{ds}=\frac{1}5 \end{align*} \right] $$ $$ \frac{1}{5}\int \frac{1}{\cos (s)}\frac{\cos (s)}{\cos (s)}\,ds= \frac{1}{5}\int \frac{\cos (s)}{\cos^2(s)}\,ds=\frac{1}{5}\int \frac{\cos (s)}{1-\sin^2(s)}\,ds $$

$$\left[ \begin{align*} \sin(s)=y\\ \frac{dy}{ds}=-\cos(s) \end{align*} \right] $$

$$ \frac{1}{5} \cdot -\frac{1}{2}\int -\frac{1}{1-y}-\frac{1}{1+y} dy $$ $$ \left[ -(-\ln(1-y) - \ln(1+y)\right] = \left[ \ln\frac{(1-y)}{(1+y)}\right] = \left[ \ln\frac{(1-\sin(s))}{(1+\sin(s))}\right] = \left[ \ln\frac{(1-\sin(5x))}{(1+\sin(5x))}\right] $$

Answer 1

Hint. Note that taking the derivative of $$\ln\left(\frac{1-\sin(5x)}{1+\sin(5x)}\right)=\ln(\underbrace{1-\sin(5x)}_{>0})-\ln(\underbrace{1+\sin(5x)}_{>0})$$ we get $$\frac{-5\cos(5x)}{1-\sin(5x)}-\frac{5\cos(5x)}{1+\sin(5x)}= -\frac{10\cos(5x)}{1-\sin^2(5x)}=-\frac{10}{\cos(5x)}.$$ So you are not so far from the correct answer: you just forgot to consider the constant $\frac{1}{5} \cdot -\frac{1}{2}$ in the last line.

Answer 2

Have you met the $\sec$ function? That will make things a lot easier here. We have that $\sec y =\frac{1}{\cos y}$, therefore your integral becomes: $$\int{\sec (5x) dx}$$Standard integration, while remembering to divide by the $5$, gives us $$\frac15\ln|\sec(5x)+\tan(5x)|+C$$

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I believe your mistake is that $$y=\sin(s)\to\frac{dy}{ds}=\cos(s) \text{ rather than } (-\cos(s))$$

Answer 3

Interestingly, you can work this out with the unintuitive subsitution $u=\sin 5x$ directly.

$$x=\frac15\arcsin u,dx=\frac15\dfrac{du}{\sqrt{1-u^2}},\\\int\frac{dx}{\cos 5x}=\frac15\int\frac{du}{\sqrt{1-u^2}\sqrt{1-u^2}}=\frac15\int\frac{du}{1-u^2}. $$

The last integral is elementary, giving

$$\frac15\int\frac{du}{1-u^2}=\frac15\text{artanh }u=\frac15\text{artanh}(\sin 5x).$$

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If you never heard of the hyperbolic functions, notice the similarity with the ordinary arc tangent, and check the logarithm-based formula. Of course, you get the same result with fraction decomposition.

Answer 4

There is nothing wrong in your answer.

Collecting your result, we indeed have

$$I=\frac15\left(-\frac12\right)\left[ \ln\frac{1-\sin(5x)}{1+\sin(5x)}\right]+C$$ which is correct as can be checked by differentiation.

It is not even necessary to consider absolute values, as the numerator and denominator are non-negative.

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Just a little reproach on you: you didn't clearly show the links between the partial results.