# Trigonometric integrals

#### paulmdrdo

##### Active member
i'm kind of unsure of my solution here please check.

$\displaystyle\int \sin^3x\cos^3x= \int(\sin x \cos x)^3 =\int(\frac{1}{2}\sin2x)^3=\frac{1}{8}\int \sin^3 2x dx$

$\displaystyle u=2x$; $\displaystyle du=2dx$; $\displaystyle dx=\frac{1}{2}du$

$\displaystyle \frac{1}{8}\int \frac{1}{2}\sin^3 u du= \frac{1}{16}\int \sin^3 udu=\frac{1}{16}\int \sin^2 u(\sin u)du=\frac{1}{16}\int \sin u(1-\cos^2 u)du$

$\displaystyle z=\cos u$; $\displaystyle dz=-\sin u du$; $\displaystyle du=-\frac{dz}{\sin u}$

$\displaystyle-\frac{1}{16}\int \sin u(1-z^2)\frac{dz}{\sin u}=-\frac{1}{16}\int (1-z^2)dz$

$\displaystyle-\frac{1}{16}\left(z-\frac{z^3}{3}\right)+C=-\frac{1}{16}z+\frac{1}{48}z^3+C$

= $\displaystyle -\frac{1}{16}\cos 2x+\frac{1}{48}\cos^3 2x+C$ ---> this is my answer

but in my book the answer is different $\displaystyle \frac{1}{4}\sin^4 x-\frac{1}{6}\sin^6 x+C$

please tell me where i was wrong.

Last edited:

#### Prove It

##### Well-known member
MHB Math Helper
Re: trigonometric integrals

Well the first thing that sticks out is that if you make the substitution \displaystyle \begin{align*} z = \cos{(u)} \end{align*} then \displaystyle \begin{align*} dz = -\sin{(u)}\,du \end{align*}...

#### paulmdrdo

##### Active member
Re: trigonometric integrals

Well the first thing that sticks out is that if you make the substitution \displaystyle \begin{align*} z = \cos{(u)} \end{align*} then \displaystyle \begin{align*} dz = -\sin{(u)}\,du \end{align*}...
i've changed my answer to this $\displaystyle -\frac{1}{16}\cos 2x+\frac{1}{48}\cos^3 2x+C$

can you help me with the other mistakes.

#### MarkFL

Staff member
Re: trigonometric integrals

I would rewrite the integral as:

$$\displaystyle \int \sin^3(x)\left(1-\sin^2(x) \right)\cos(x)\,dx=\int\sin^3(x)\cos(x)\,dx-\int\sin^5(x)\cos(x)\,dx$$

and then for both integrals use the substitution:

$$\displaystyle u=\sin(x)\,\therefore\,du=\cos(x)\,dx$$

#### SuperSonic4

##### Well-known member
MHB Math Helper
Re: trigonometric integrals

= $\displaystyle -\frac{1}{16}\cos 2x+\frac{1}{48}\cos^3 2x+C$ ---> this is my answer

but in my book the answer is different $\displaystyle \frac{1}{4}\sin^4 x-\frac{1}{6}\sin^6 x+C$

please tell me where i was wrong.
I get the same result following your work so I feel the book is a different version of yours.

$\cos(2x) = 1-2\sin^2(x)$

$\displaystyle -\frac{1}{16}(1-2\sin^2(x))+\frac{1}{48}(1-2\sin^2(x))^3+C$

$\displaystyle -\frac{1}{16} + \frac{1}{8}\sin^2(x)) +\frac{1}{48} ( 1 - 6\sin^2(x) + 12\sin^4(x) - 8\sin^6(x))+ C$

$\displaystyle -\frac{1}{16} + \frac{1}{8}\sin^2(x)) +\frac{1}{48} - \frac{1}{8}\sin^2(x) + \frac{1}{4}\sin^4(x) - \frac{1}{6}\sin^6(x)+ C$

$\displaystyle -\frac{1}{16} +\frac{1}{48} + \frac{1}{4}\sin^4(x) - \frac{1}{6}\sin^6(x)+ C$

$\displaystyle -\frac{1}{16} +\frac{1}{48} + \frac{1}{4}\sin^4(x) - \frac{1}{6}\sin^6(x)+ C' \text{ where } C' = C + \frac{1}{16} - \frac{1}{48}$

$\displaystyle \frac{1}{4}\sin^4(x) - \frac{1}{6}\sin^6(x)+ C'$