How do you find the antiderivative of $\int {sin}^{3} x cos x d x$ $int sin^3xcosxdx$ ?

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How do you find the antiderivative of $\int {sin}^{3} x cos x d x$ $int sin^3xcosxdx$ ?

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Answer 1 · 2016-11-23T17:23:09

$\int {sin}^{3} x cos x d x = \frac{1}{4} {sin}^{4} x + C$ $" "intsin^3xcosxdx" "=1/4sin^4x+C$

Explanation:

no need for substitution here if you recognise that

$y = {sin}^{n} x \Rightarrow \frac{d y}{d x} = n {sin}^{n - 1} x cos x$ $y=sin^nx=>(dy)/(dx)=nsin^(n-1)xcosx" "$ using the chain rule

so $\int {sin}^{3} x cos x d x$ $" "intsin^3xcosxdx" "$ suggests a function of the type

$y = {sin}^{4} x$ $y=sin^4x$

lets check this by differentiating.

$u = sin x \Rightarrow \frac{d u}{d x} = cos x$ $u=sinx=>(du)/(dx)=cosx$

$y = u^{4} \Rightarrow \frac{d y}{d x} = 4 u^{3}$ $y=u^4=>(dy)/(dx)=4u^3$

$\frac{d y}{d x} = 4 u^{3} cos x = 4 {sin}^{3} x cos x$ $(dy)/(dx)=4u^3cosx=4sin^3xcosx$

$\int {sin}^{3} x cos x d x = \frac{1}{4} {sin}^{4} x + C$ $" "intsin^3xcosxdx" "=1/4sin^4x+C$

Answer 2 · 2016-11-23T17:36:49

$\frac{{sin}^{4} x}{4} + C$ $sin^4x/4+C$ .

Explanation:

Since you have a cosine terms hanging around some sine terms, it might be helpful to try the substitution $u = sin x$ $u=sinx$ , $d u = cos x d x$ $du=cosxdx$ .

Using this substitution, $\int {sin}^{3} x cos x d x = \int u^{3} d u$ $intsin^3xcosxdx=intu^3du$ .

$\int u^{3} d u = \frac{u^{4}}{4} + C = \frac{{sin}^{4} x}{4} + C$ $intu^3du=u^4/4+C=sin^4x/4+C$ .

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How do you find the antiderivative of $\int {sin}^{3} x cos x d x$ $int sin^3xcosxdx$ ?

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