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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:

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

given by:

https://www.symbolab.com/solver/step-by-step/%5Cint%20cos%5E%7B5%7D%5Cleft%28x%5Cright%29dx/?origin=button

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    $\begingroup$ Use $sin^{2}x = 1 - cos^{2}x$ $\endgroup$
    – openspace
    Apr 10, 2016 at 20:03
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    $\begingroup$ 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

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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:

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

$=\frac{1}{5}\sin(x)-\frac{2}{5}\sin^3(x)+\frac{1}{5}\sin^5(x)+\frac{4}{15}\sin(x)-\frac{4}{15}\sin^3(x))+\frac{8}{15}\sin(x)$

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

Which agrees with the answer from the solver.

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