How do I evaluate the indefinite integral $\int {cot}^{5} (x) \cdot {sin}^{4} (x) d x$ $intcot^5(x)*sin^4(x)dx$ ?

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How do I evaluate the indefinite integral $\int {cot}^{5} (x) \cdot {sin}^{4} (x) d x$ $intcot^5(x)*sin^4(x)dx$ ?

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Answer 1 · 2018-04-23T21:31:25

${cos}^{2} x + \frac{1}{4} {sin}^{4} x + ln | sin x | + c$ $cos^2x+1/4sin^4x+lnabssinx+"c"$

Explanation:

${cot}^{5} x {sin}^{4} x = \frac{{cos}^{5} x}{{sin}^{5} x} {sin}^{4} x = \frac{{cos}^{5} x}{sin x} = {cos}^{4} x cot x = {(1 - {sin}^{2} x)}^{2} cot x = (1 - 2 {sin}^{2} x + {sin}^{4} x) cot = cot x - 2 sin x cos x + {sin}^{3} x cos x$ $cot^5xsin^4x=cos^5x/sin^5xsin^4x=cos^5x/sinx=cos^4xcotx=(1-sin^2x)^2cotx=(1-2sin^2x+sin^4x)cot=cotx-2sinxcosx+sin^3xcosx$

So

$\int {cot}^{5} x {sin}^{4} x d x = \int cot x d x - \int 2 sin x cos x d x + \int {sin}^{3} x cos x d x$ $intcot^5xsin^4xdx=intcotxdx-int2sinxcosxdx+intsin^3xcosxdx$

Now let $u = sin x$ $u=sinx$ and $d u = cos x d x$ $du=cosxdx$ and $v = cos x$ $v=cosx$ and $d v = - sin x d x$ $dv=-sinxdx$

$\int cot x d x + \int - 2 sin x cos x d x + \int {sin}^{3} x cos x d x = ln | sin x | \int 2 v d v + \int u^{3} d u = ln | sin x | + v^{2} + \frac{1}{4} u^{4} + c = {cos}^{2} x + \frac{1}{4} {sin}^{4} x + ln | sin x | + c$ $intcotxdx+int-2sinxcosxdx+intsin^3xcosxdx=lnabssinx int2vdv +intu^3du=lnabssinx+v^2+1/4u^4+"c"=cos^2x+1/4sin^4x+lnabssinx+"c"$

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How do I evaluate the indefinite integral $\int {cot}^{5} (x) \cdot {sin}^{4} (x) d x$ $intcot^5(x)*sin^4(x)dx$ ?

1 Answer

Explanation:

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How do I evaluate the indefinite integral $\int {cot}^{5} (x) \cdot {sin}^{4} (x) d x$ $intcot^5(x)*sin^4(x)dx$ ?