What is $\int {cos}^{3} (x) {sin}^{4} (x) d x$ $int cos^3(x) sin^4(x) dx$ ?

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Answer 1 · 2018-04-27T01:07:52

The integral is equal to $\frac{7 {sin}^{5} x - 5 {sin}^{7} x}{35} + C$ $(7sin^5x-5sin^7x)/35+C$ .

Use ${cos}^{2} x = 1 - {sin}^{2} x$ $cos^2x=1-sin^2x$ , then use a substitution:

$= \int {cos}^{3} x \cdot {sin}^{4} x$ $color(white)=intcos^3x*sin^4x$ $d x$ $dx$

$= \int cos x \cdot {cos}^{2} x \cdot {sin}^{4} x$ $=intcosx*cos^2x*sin^4x$ $d x$ $dx$

$= \int cos x \cdot (1 - {sin}^{2} x) \cdot {sin}^{4} x$ $=intcosx*(1-sin^2x)*sin^4x$ $d x$ $dx$

Let $u = sin x$ $u=sinx$ , so $d u = cos x$ $du=cosx$ $d x$ $dx$ , and $d x = \frac{d u}{cos x}$ $dx=(du)/cosx$ :

$= \int cos x \cdot (1 - u^{2}) \cdot u^{4} \cdot \frac{d u}{cos x}$ $=intcosx*(1-u^2)*u^4*(du)/cosx$

$=intcolor(red)cancelcolor(black)cosx*(1-u^2)*u^4*(du)/color(red)cancelcolor(black)cosx$

$=int(1-u^2)*u^4$ $du$

$=int(u^4-u^6)$ $du$

$=intu^4$ $du-intu^6$ $du$

$=u^5/5-u^7/7+C$

$=(7u^5-5u^7)/35+C$

$=(7sin^5x-5sin^7x)/35+C$

That's the integral. Hope this helped!

What is int cos^3(x) sin^4(x) dx ∫cos3(x)sin4(x)dxint cos^3(x) sin^4(x) dx ?