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I've looked on the web for an answer to this question, and could not find an example.

Could you push me towards a proper u substitution for the following integral? Please don't solve the problem just state what you would use as a substitution and why.

$$\int(\sin^{10}x \cdot \cos x)\ dx$$

My sad attempt

let $$u=\sin x$$ $$du=\cos{x}\ dx$$

$$\int(\sin^{9}x) du$$

should I use trigonometric identities or is another substitution valid?

This question is in the substitution section of the textbook, so it has to be solved with simple substitution

Thanks.

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When you substitute $u=\sin x$ you should carry it through, replacing every occurrence of $\sin x$ with $u.$ That reduces the problem to integrating a simple power.

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The substution works because $u=\sin x \implies \frac{du}{dx}= \cos x $

You are the using the equivalent operators $\color {red}{\int}u^{10} \color {red}{ \frac {du}{dx}dx} \equiv \color {red}{\int} u^{10} \color{red}{du}$

Notice how the $\cos x$ is swallowed up.

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If you need a spoiler:

$$\int u^{10}du=\frac{u^{11}}{11}+C\\=\frac{\sin^{11}(x)}{11}+C$$

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  • $\begingroup$ They said please don't solve the problem... $\endgroup$ – Justpassingby Feb 1 '16 at 19:32
  • $\begingroup$ @Justpassingby sorry $\endgroup$ – 3SAT Feb 1 '16 at 19:33
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thanks to u/justpassingthrough

$$\int(sin^{10}x*cosx)dx$$ let $$u=sinx$$ $$du=cosx$$

$$\int(sin^{10}x*cosx)dx=\int(u^{10})du$$ $$=\frac{1}{11}u^{11}+C=\frac{1}{11}(sinx)^{11}+C$$

cool

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