# Math Help - Trigonometric integral....

1. ## Trigonometric integral....

any help with this?

$\int^{\pi/3}_{0} \frac{\sin^3x}{\sqrt{\cos x}} \, dx$

I first divided the $\sin^3 x$ into two: $\frac{\sin^2 x \sin x}{ \sqrt{\cos x}}$

then got $\frac{\sin^2 x \sin x}{ \cos^{1/2} x}$

I get confused here but I think it is....

(-cos^2x) sinx cos^1/2x

i get stuck here....

2. Originally Posted by dude487
any help with this?

$
\int\frac{sin^3x}{cosx}

$
Hello,

I fixed the latex. should be { } not ( )

have you tried substituting $t=\sin(x)$ ?
$dt=\cos(x) ~dx$

so we have $\int\frac{t^3}{\cos^2(x)} ~dt$
and $\cos^2(x)=1-\sin^2(x)=1-t^2$

which will give a partial decomposition...

so let's try $u=\cos(x)$
$du=-\sin(x) ~dx$

so we have $\int \frac{-\sin^2(x)}{u} ~du=\int\frac{u^2-1}{u} ~du=\int u-\frac 1u ~du$

much better ! can you finish it ?

--------------------------------------------------------
okay, you changed the problem... however, it doesn't change the method. the best substitution seems to be u=cos(x)
as for the boundaries, i'll leave it to you =)

3. Originally Posted by dude487
any help with this?

$\int^{\pi/3}_{0} \frac{\sin^3x}{\sqrt{\cos x}} \, dx$

I first divided the $\sin^3 x$ into two: $\frac{\sin^2 x \sin x}{ \sqrt{\cos x}}$

then got $\frac{\sin^2 x \sin x}{ \cos^{1/2} x}$

I get confused here but I think it is....

(-cos^2x) sinx cos^1/2x

i get stuck here....
$\int_0^{\pi /3} {\frac{{{{\sin }^3}x}}{{\sqrt {\cos x} }}dx} = \int_0^{\pi /3} {\frac{{\sin x{{\sin }^2}x}}{{\sqrt {\cos x} }}dx} = \int_0^{\pi /3} {\frac{{\sin x\left( {1 - {{\cos }^2}x} \right)}}{{\sqrt {\cos x} }}dx} =$

$= \int_0^{\pi /3} {\sin x\left( {\frac{1}{{\sqrt {\cos x} }} - \sqrt {{{\cos }^3}x} } \right)dx} = \int_0^{\pi /3} {\left( {\sqrt {{{\cos }^3}x} - \frac{1}{{\sqrt {\cos x} }}} \right)d\left( {\cos x} \right)} =$

$= \left. {\left( {\frac{2}{5}\sqrt {{{\cos }^5}x} - 2\sqrt {\cos x} } \right)} \right|_0^{\pi /3} = \left. {2\sqrt {\cos x} \left( {\frac{1}
{5}{{\cos }^2}x - 1} \right)} \right|_0^{\pi /3} =$

$= 2 \cdot \frac{1}{{\sqrt 2 }}\left( {\frac{1}{5} \cdot \frac{1}{4} - 1} \right) - 2\left( {\frac{1}{5} - 1} \right) = \sqrt 2 \left( {\frac{1}
{{20}} - 1} \right) - 2\left( {\frac{1}{5} - 1} \right) = \frac{8}
{5} - \frac{{19\sqrt 2 }}{{20}}.$