How do you integrate $\int \frac{sin x}{{(cos x)}^{5}}$ $int sinx/(cosx)^5$ using substitution?

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Answer 1 · 2018-06-16T11:38:29

The answer is $= \frac{1}{4 {cos}^{4} x} + C$ $=1/(4cos^4x)+C$

Let $u = cos x$ $u=cosx$

$\Rightarrow$ $=>$ , $d u = - sin x d x$ $du=-sinxdx$

The integral is

$I = \int \frac{sin x d x}{{cos}^{5} x} = - \int \frac{d u}{u^{5}}$ $I=int(sinxdx)/(cos^5x)=-int(du)/u^5$

$= \frac{1}{4 u^{4}}$ $=1/(4u^4)$

$= \frac{1}{4 {cos}^{4} x} + C$ $=1/(4cos^4x)+C$

How do you integrate int sinx/(cosx)^5∫sinx(cosx)5int sinx/(cosx)^5 using substitution?