I'm trying to calculate $$\int cos^5(x)dx$$ with the reduction formula.

$$\int cos^5(x)dx=\frac{1}{5}cos^4(x)sin(x)+\frac{4}{5}\int cos^3(x)dx$$ then $$\int cos^3(x)dx=\frac{1}{3}cos^2(x)sin(x)+\frac{2}{3}\int cos(x)dx$$ so $$\int cos^5(x)dx=\frac{1}{5}cos^4(x)sin(x)+\frac{4}{5}(\frac{1}{3}cos^2(x)sin(x)+\frac{2}{3}sin(x))$$

But this looks different from:


given by:


  • 2
    $\begingroup$ Use $sin^{2}x = 1 - cos^{2}x$ $\endgroup$
    – openspace
    Apr 10, 2016 at 20:03
  • 1
    $\begingroup$ This is typically trigonometry. You arrive at seemingly different answers and yet, they differ by only a constant. One thing you can do before you dive into a trigo-algebraic maze of computations, is to graph both of your answers to see if the graphs differ by a constant. If so, it is worth to do the algebra. If not, well....there is likely to be some mistake. $\endgroup$
    – imranfat
    Apr 10, 2016 at 20:12

1 Answer 1


Converting the cosines in your answer to sines via the pythagorean identity, $\cos ^2x=1-\sin ^2x$ yields:

$\frac{1}{5}\cos^4(x)\sin(x)+\frac{4}{5}(\frac{1}{3}\cos^2(x)\sin(x)+\frac{2}{3}\sin(x))$ $=\frac{1}{5}(1-\sin^2(x))^2\sin(x)+\frac{4}{5}(\frac{1}{3}(1-\sin^2(x))\sin(x)+\frac{2}{3}\sin(x))$

Simplifying from there:




Which agrees with the answer from the solver.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .