5
$\begingroup$

I tried using trigonometric formulas for turning it into 2$\int \frac{\cos^2(2x)}{\cos(x)}dx - \int \frac{1}{\cos(x)}dx$ and can solve the second one, but still no idea of how to proceed with $\int \frac{\cos^2(2x)}{\cos(x)}dx$. Any suggestion?

$\endgroup$
6
$\begingroup$

I would suggest that you instead use the cosine’s quadruple-angle formula (which can be derived by applying the double-angle formula twice):

$$\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$

which would turn your integral into

$$\int \frac{\cos(4x)}{\cos(x)}dx=8\int \cos^3(x)dx-8\int \cos(x)dx+\int \sec(x)dx$$

and you probably know how to evaluate each of these integrals.

$\endgroup$
4
$\begingroup$

Note that$$\cos^2(2x)=\bigl(\cos^2(x)-\sin^2(x)\bigr)^2=\bigl(2\cos^2(x)-1\bigr)^2=4\cos^4(x)-4\cos^2(x)+1$$and that therefore$$\int\frac{\cos^2(2x)}{\cos(x)}\,\mathrm dx=4\int\cos^3(x)\,\mathrm dx-4\int\cos(x)\,\mathrm dx+\int\frac1{\cos(x)}\,\mathrm dx.$$

$\endgroup$
4
$\begingroup$

Hint:

By Werner Formulas

$$\cos(n+2)x+\cos nx=2\cos x\cos(n+1)x$$

For $\cos x\ne0,$

$$\implies\int\dfrac{\cos(n+2)x}{\cos x}\ dx=2\int\cos(n+1)x\ dx-\int\dfrac{\cos nx}{\cos x}\ dx$$

Set $n=2$ and then $n=0$

Finally use Integral of the secant function

$\endgroup$
1
  • $\begingroup$ This is too an interesting one! $\endgroup$ Jan 19 '19 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.