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I tried substituting $\cos(x) = 1 - 2\sin^2(x/2)$ but still can't figure it out. Is there any other identity to help with this integration?

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closed as off-topic by Eevee Trainer, YuiTo Cheng, Brevan Ellefsen, Jean-Claude Arbaut, Shailesh May 14 at 6:11

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  • 5
    $\begingroup$ $1+\cos x=2\cos^2\frac{x}{2}$. $\endgroup$ – Nosrati May 8 at 6:54
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You can use $\cos(x)=2\cos^2(\frac{x}{2})-1$ the next step can be to use $\cos^5(\frac{x}{2})=\cos(\frac{x}{2})(1-\sin^4(\frac{x}{2}))$

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  • 3
    $\begingroup$ Just remember about the absolute value: $\big(\cos^2\frac{x}{2}\big)^\frac52 = |\cos \frac{x}{2}|^5$ $\endgroup$ – Adam Latosiński May 8 at 7:08
  • $\begingroup$ Okay. Just a query does it matter in an indefinite integral. $\endgroup$ – Archis Welankar May 8 at 7:58
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    $\begingroup$ Yes it does; If you want to write the formula for the indefinite integral on the domain $D=\mathbb{R}$, that means that for some $x$ its derivative is supposed to be $(\cos \frac{x}{2})^5$ and for some other it is supposed to be $-(\cos \frac{x}{2})^5$. If you neglect the absolute value, you won't get that result. $\endgroup$ – Adam Latosiński May 8 at 8:38
  • $\begingroup$ Thanks for answering. $\endgroup$ – Archis Welankar May 8 at 13:06

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