Question

Integral: Sin^3 (x)

Answer 1

I dont know how to solve \[\int\limits_{}^{} \sin ^{3}x\]

Answer 2

\[\int sin^{n}(x)dx=-\frac{sin^{n-1}(x)\ cos(x)}{n}+\frac{n-1}{n}\int sin^{n-2}(x)dx\]

Answer 3

\[\sin^2x * \sin x\]
\[(1 - \cos ^2 x )*\sin x\]

Answer 4

there are a few different reduction formulas; the most general is the one I posted

Answer 5

Hmm oke, but I also saw this reduction

\[\sin ^{3}x = \sin x (1-\cos ^{2}x)\]

could you explain how to intergrate this reduction?

Answer 6

\[\int sin^{n}(x)dx=-\frac{sin^{n-1}(x)\ cos(x)}{n}+\frac{n-1}{n}\int sin^{n-2}(x)dx\]
\[\int cos^{n}(x)dx=\frac{cos^{n-1}(x)\ sin(x)}{n}+\frac{n-1}{n}\int cos^{n-2}(x)dx\]
\[\int tan^{n}(x)dx=\frac{1}{n}tan^{n-1}(x)-\int tan^{n-2}(x)dx\]

Answer 7

\[\int\sin{x} - \int \sin{x}*\cos^2{x}\]Use \(u=\cos{x}\) Substitution for the second one

Answer 8

wow; thank you @amistre

Answer 9

while ishaan does that, ill practice the general one :)

\[\int sin^{3}(x)dx=-\frac{sin^{2}(x)\ cos(x)}{3}+\frac{2}{3}\int sin(x)dx\]
\[=-\frac{sin^{2}(x)\ cos(x)}{3}-\frac{2}{3}cos(x)\]
\[=-\frac{1}{3}cos(x)(2+sin^{2}(x))\]
which turns out to be another reduction formula :)

Answer 10

Ishaan i have to integrate by parts.

u = cos x
du = -sin x

\[\int\limits_{}^{}\sin x - \int\limits_{}^{}sinx*cosx\]
\[-\cos x + \int\limits_{}^{}u ^{2} du\]

\[-cosx + (1/3)\cos ^{3}x\]?????

Answer 11

Yea, you did that right....

Answer 12

but how can that be

\[-(1/3)cosx(2+\sin ^{2}x)\]?

Answer 13

my question isnt answered yet:S

Answer 14

where s your confusion at?

Answer 15

as i posted my solution is \[-cosx+(1/3)\cos ^{3}x\]

but the answer is:

\[-(1/3)cosx(2+\sin ^{2}x)\]

Answer 16

if anything, it in the algebra and the trig identities

Answer 17

2+sin^2x = 2 + 1 -cos^2x = 3 - cos^2x
-(1/3)(cosx)(3-cos^2x) = -cosx +(1/3)cos^3x
which is your answer

Answer 18

Ahh oke.. thank you :) i got it now :)

Answer 19

i think i leave it as it is ^^

Answer 20

\begin{align}
\int sin^3(x)dx&=\int sin^2(x)sin(x)dx\\\\
&=\int (1-cos^2(x))sin(x)dx\\\\
&=\int sin(x)-cos^2(x)sin(x)dx\\\\
&=\int sin(x)dx-\int cos^2(x)sin(x)dx\\\\
&=-cos(x)+\frac{1}{3} cos^3(x)\\\\
&=-cos(x)(1+\frac{1}{3} cos^2(x))\\\\
&=-cos(x)(\frac{3+cos^2(x)}{3} )\\\\
&=-\frac{1}{3}cos(x)(3+cos^2(x) )\\\\
&=-\frac{1}{3}cos(x)(3+(1-sin^2(x))\\\\
\end{align}

well, its getting there but I messes up in the algebra someplace lol

Answer 21

hahaahahha.... thank you for the solution :)

Answer 22

<3 thank you for your time ;)

Answer 23

i see it, when I pulled out the -cos(x) i forgot to change out the sign inside ....its spose to be:
\[-cos(1-\frac{1}{3}cos^2(x))\]
\[-\frac{1}{3}cos(3-cos^2(x))\]
\[-\frac{1}{3}cos(3-(1-sin^2(x))\]
\[-\frac{1}{3}cos(2+sin^2(x))\]
tadaa!! ;)