Two indefinite integrals, the advanced section

**Dext91** · July 18th 2009, 06:58 AM

Can anyone help me to solve these two indefinite integrals?

1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
2) Find int [cos(x)/(sin(x+1))^2]dx.

For me this integrals were too complicated.
Thanks in advance!

**pomp** · July 18th 2009, 07:14 AM

Originally Posted by Dext91

1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].

Substitue $u = 1+\cos{x}$ and you arrive at $\int \frac{du}{u^{3/2} (u-2)}$ .

Now use the subtitution $v = \sqrt{u}$ . It should be quite managable from here on, you might want to consider using partial fractions for the next step.

Good luck, pomp.

**VonNemo19** · July 18th 2009, 07:22 AM

Originally Posted by Dext91

1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].

$ \int{\frac{dx}{\sin{x}\cdot\sqrt{1+\cos{x}}}} $

Rewrite:
$ \int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx $

Let $u=1+\cos{x}$ , then $du=-\sin{x}dx$

Therefore $\int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx=-\int{\frac{1}{\sqrt{u}}}du$

Not sure if this is correct.

**pomp** · July 18th 2009, 07:46 AM

Originally Posted by VonNemo19

$ \int{\frac{dx}{\sin{x}\cdot\sqrt{1+\cos{x}}}} $

Rewrite:
$ \int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx $

Let $u=1+\cos{x}$ , then $du=-\sin{x}dx$

Therefore $\int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx=-\int{\frac{1}{\sqrt{u}}}du$

Not sure if this is correct.

Slightly wrong, you multiplied through by sin instead of dividing when you subbed in.

$dx = - \frac{du}{\sin{x}}$ . Subbing in gives:

$\int \frac{\frac{-du}{\sin{x}}}{\sin{x} \sqrt{u}} = \int - \frac{du}{\sin^2{x} \sqrt{u}}$

$\sin^2{x} = 1 - (u-1)^2$

So the integral becomes $\int \frac{ -du}{(2u - u^2)\sqrt{u}}$

**galactus** · July 18th 2009, 07:54 AM

Is that second one supposed to be:

$\int\frac{cos(x)}{sin^{2}(x+1)}dx$

or

$\int\frac{cos(x)}{(sin(x)+1)^{2}}dx$

I am thinking the latter since that would be easier. Do you have a typo?. Just checking.

**pomp** · July 18th 2009, 07:56 AM

Originally Posted by galactus

Is that second one supposed to be:

$\int\frac{cos(x)}{sin^{2}(x+1)}dx$

or

$\int\frac{cos(x)}{(sin(x)+1)^{2}}dx$

I am thinking the latter since that would be easier. Do you have a typo?. Just checking.

I hope you're right. After putting the former through matlab I'm afraid to even attempt it.

**galactus** · July 18th 2009, 07:59 AM

Yes indeed, Pomp. If it is the latter, the simple sub $u=sin(x), \;\ du=cos(x)$ will suffice. But the former is a booger.

**Calculus26** · July 18th 2009, 08:01 AM

Not quite the sin(x) term does not cancel but rather

you have 1/[sin^2(x)sqrt(u)]

See attachment we can avoid sqr roots altogether

**DeMath** · July 18th 2009, 08:28 AM

Originally Posted by Dext91

Can anyone help me to solve these two indefinite integrals?

1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
2) Find int [cos(x)/(sin(x+1))^2]dx.

For me this integrals were too complicated.
Thanks in advance!

For number one

$\int {\frac{{dx}} {{\sin x\sqrt {1 + \cos x} }}} = \int {\frac{{\sin xdx}} {{{{\sin }^2}x\sqrt {1 + \cos x} }}} = \int {\frac{{\sin xdx}} {{\left( {1 - {{\cos }^2}x} \right)\sqrt {1 + \cos x} }}} =$

$= \left\{ \begin{gathered}\cos x = t, \hfill \\\sin xdx = - dt \hfill \\ \end{gathered} \right\}$ $= - \int {\frac{{dt}}{{\left( {1 - {t^2}} \right)\sqrt {1 + t} }}} = - \frac{1}{2}\int {\frac{{1 - t + 1 + t}}{{\left( {1 - t} \right)\left( {1 + t} \right)\sqrt {1 + t} }}dt} =$

$= - \frac{1}{2}\int {\frac{{dt}}{{\left( {1 + t} \right)\sqrt {1 + t} }}} - \frac{1}{2}\int {\frac{{dt}}{{\left( {1 - t} \right)\sqrt {1 + t} }}} = {I_1} - {I_2}.$

${I_1} = - \frac{1}{2}\int {\frac{{dt}}{{\sqrt {{{\left( {1 + t} \right)}^3}} }}} = \frac{1}{{\sqrt {1 + t} }} + C = \frac{1}{{\sqrt {1 + \cos x} }} + C.$

${I_2} = \frac{1}{2}\int {\frac{{dt}}{{\left( {1 - t} \right)\sqrt {1 + t} }}} = \frac{1}{2}\int {\frac{{{1 \mathord{\left/{\vphantom {1 {\sqrt {1 + t} }}} \right.\kern-\nulldelimiterspace} {\sqrt {1 + t} }}}} {{2 - {{\left( {\sqrt {1 + t} } \right)}^2}}}dt}$ $= \frac{1}{4}\int {\frac{{{1 \mathord{\left/{\vphantom {1 {\sqrt {1 + t} }}} \right.\kern-\nulldelimiterspace} {\sqrt {1 + t} }}}}{{1 - {{\left( {\frac{{\sqrt 2 }}{2}\sqrt {1 + t} } \right)}^2}}}} dt =$

$= \left\{ \begin{gathered}\frac{{\sqrt 2 }}{2}\sqrt {1 + t} = u, \hfill \\\frac{{dt}}{{4\sqrt {1 + t} }} = \frac{{\sqrt 2 }}{2}du \hfill \\ \end{gathered} \right\} =$ $\frac{{\sqrt 2 }}{2}\int {\frac{{du}}{{1 - {u^2}}}} = \frac{{\sqrt 2 }}{2}\int {\frac{{du}}{{\left( {1 - u} \right)\left( {1 + u} \right)}}} =$

$= \frac{{\sqrt 2 }}{4}\int {\left( {\frac{1}{{1 + u}} + \frac{1} {{1 - u}}} \right)du} = \frac{{\sqrt 2 }}{4}\left( {\ln \left| {1 + u} \right| - \ln \left| {1 - u} \right|} \right) + C =$

$= \frac{{\sqrt 2 }}{4}\left( {\ln \left| {1 + \frac{{\sqrt 2 }} {2}\sqrt {1 + t} } \right| - \ln \left| {1 - \frac{{\sqrt 2 }} {2}\sqrt {1 + t} } \right|} \right) + C =$

$= \frac{{\sqrt 2 }} {4}\left( {\ln \left| {\sqrt 2 + \sqrt {1 + t} } \right| - \ln \left| {\sqrt 2 - \sqrt {1 + t} } \right|} \right) + C =$

$= -\frac{{\sqrt 2 }}{4}\ln \left| {\frac{{\sqrt 2 - \sqrt {1 + t} }}{{\sqrt 2 + \sqrt {1 + t} }}} \right| + C = -\frac{{\sqrt 2 }}{4}\ln \left| {\frac{{\sqrt 2 - \sqrt {1 + \cos x} }}{{\sqrt 2 + \sqrt {1 + \cos x} }}} \right| + C.$

Finally

$\int {\frac{{dx}}{{\sin x\sqrt {1 + \cos x} }}} = \frac{1} {{\sqrt {1 + \cos x} }} + \frac{{\sqrt 2 }}{4}\ln \left| {\frac{{\sqrt 2 - \sqrt {1 + \cos x} }}{{\sqrt 2 + \sqrt {1 + \cos x} }}} \right| + C.$

**DeMath** · July 18th 2009, 08:59 AM

Originally Posted by Dext91

Can anyone help me to solve these two indefinite integrals?

1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
2) Find int [cos(x)/(sin(x+1))^2]dx.

For me this integrals were too complicated.
Thanks in advance!

For number 2

$\int {\frac{{\cos \left( x \right)}}{{{{\sin }^2}\left( {x + 1} \right)}}} dx = \int {\frac{{\cos \left( {x + 1 - 1} \right)}}{{{{\sin }^2}\left( {x + 1} \right)}}} dx =$

$= \int {\frac{{\cos \left( {x + 1} \right)\cos \left( 1 \right) + \sin \left( {x + 1} \right)\sin \left( 1 \right)}}{{{{\sin }^2}\left( {x + 1} \right)}}} dx =$

$= \cos \left( 1 \right)\int {\frac{{\cos \left( {x + 1} \right)}} {{{{\sin }^2}\left( {x + 1} \right)}}} dx + \sin \left( 1 \right)\int {\frac{{dx}}{{\sin \left( {x + 1} \right)}}} =$

$= \cos \left( 1 \right)\int {\frac{{d\left( {\sin \left( {x + 1} \right)} \right)}}{{{{\sin }^2}\left( {x + 1} \right)}}} + \frac{{\sin \left( 1 \right)}}{2}\int {\frac{{dx}}{{\sin \left( {\frac{{x + 1}}{2}} \right)\cos \left( {\frac{{x + 1}}{2}} \right)}}} =$

$= - \frac{{\cos \left( 1 \right)}}{{\sin \left( {x + 1} \right)}} + \frac{{\sin \left( 1 \right)}}{2}\int {\frac{{dx}}{{{{\cos }^2}\left( {\frac{{x + 1}}{2}} \right){\text{tg}}\left( {\frac{{x + 1}}{2}} \right)}}} =$

$= - \frac{{\cos \left( 1 \right)}}{{\sin \left( {x + 1} \right)}} + \sin \left( 1 \right)\int {\frac{{d\left[ {{\text{tg}}\left( {\frac{{x + 1}}{2}} \right)} \right]}}{{{\text{tg}}\left( {\frac{{x + 1}}{2}} \right)}}} =$

$= \sin \left( 1 \right)\ln \left[ {{\text{tg}}\left( {\frac{{x + 1}} {2}} \right)} \right] - \frac{{\cos \left( 1 \right)}}{{\sin \left( {x + 1} \right)}} + C.$

**Calculus26** · July 18th 2009, 09:03 AM

My original attachment contains a mistake--see correct version

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