Hard Integrals

Ment · #1 $Old$ January 3rd, 2009, 10:45 AM

Please, help me solution these integrals:

$1.{\text{ }}\int {\frac{{\sin x\cos x}} {{\sin ^4 x + \cos ^4 x}}dx} .$

$2.{\text{ }}\int {\frac{{\arctan e^x }}{{\cosh x}}} dx.$

$3.{\text{ }}\int {\frac{{\cos x}}{{\sqrt {1 + \sin x + \cos ^2 x} }}dx} .$

$4.{\text{ }}\int {\frac{{dx}}{{a + b\cos x}}} {\text{ if }}1){\text{ }}a > 0{\text{ }} \wedge {\text{ }}b > 0;{\text{ }}2){\text{ }}a > 0{\text{ }} \wedge {\text{ }}b < 0.$

Chop Suey · #2 $Old$ January 3rd, 2009, 10:55 AM

2. Recall that $\cosh{x} = \frac{e^x + e^{-x}}{2}$
Apply this identity and multiply by $\frac{e^x}{e^x}$ and you'll get a simple integral.

3. Use the identity $\cos^2{x} = 1 - \sin^2{x}$ and complete the square. Again, simple integral.

I will return later for 1 and 4, if no one beats me to it.

Chop Suey · #3 $Old$ January 3rd, 2009, 10:59 AM

For 1, multiply by $\frac{\displaystyle\frac{1}{\cos^4{x}}}{\displaystyle\frac{1}{\cos^4{x}}}$ and you'll get a simple integral.

Danny · #4 $Old$ January 3rd, 2009, 10:59 AM

Quote:

Originally Posted by Ment $View Post$

Please, help me solution these integrals:

$1.{\text{ }}\int {\frac{{\sin x\cos x}}{{\sin ^4 x + \cos ^4 x}}dx} .$

$2.{\text{ }}\int {\frac{{\arctan e^x }}{{\cosh x}}} dx.$

$3.{\text{ }}\int {\frac{{\cos x}}{{\sqrt {1 + \sin x + \cos ^2 x} }}dx} .$

$4.{\text{ }}\int {\frac{{dx}}{{a + b\cos x}}} {\text{ if }}1){\text{ }}a > 0{\text{ }} \wedge {\text{ }}b > 0;{\text{ }}2){\text{ }}a > 0{\text{ }} \wedge {\text{ }}b < 0.$

Some hints:
(1) try using $\sin^2x + \cos^2 x = 1$

(2) try using $u = e^x$

(1) try using $u = \sin x$ and use a trig identity

(1) try using $x = 2 \tan^{-1} u$

Danny · #5 $Old$ January 3rd, 2009, 11:00 AM

Pretty fast Chop Suey

Chop Suey · #6 $Old$ January 3rd, 2009, 11:34 AM

Quote:

Originally Posted by danny arrigo $View Post$

Some hints:
(1) try using $x = 2 \tan^{-1} u$

Quote:

Originally Posted by danny arrigo $View Post$

Pretty fast Chop Suey

Not fast enough

DeMath · #7 $Old$ January 3rd, 2009, 03:00 PM

1. $\int {\frac{{\sin x\cos x}}{{\sin ^4 x + \cos ^4 x}}dx} .$

$\begin{gathered}\sin ^4 x + \cos ^4 x = \sin ^4 x + 2\sin ^2 x\cos ^2 x + \cos ^4 x - 2\sin ^2 x\cos ^2 x = \hfill \\ = 1 - 2\left( {\sin x\cos x} \right)^2 . \hfill \\ \end{gathered}$

$\int {\frac{{\sin x\cos x}}{{\sin ^4 x + \cos ^4 x}}dx} = \int {\frac{{\sin x\cos x}}{{1 - 2\left( {\sin x\cos x} \right)^2 }}dx = }$

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

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

Danny · #8 $Old$ January 3rd, 2009, 03:35 PM

Quote:

Originally Posted by danny arrigo $View Post$

Some hints:
(1) try using $\sin^2x + \cos^2 x = 1$

Quote:

Originally Posted by DeMath $View Post$

1. $\int {\frac{{\sin x\cos x}}{{\sin ^4 x + \cos ^4 x}}dx} .$

$\begin{gathered}\sin ^4 x + \cos ^4 x = \sin ^4 x + 2\sin ^2 x\cos ^2 x + \cos ^4 x - 2\sin ^2 x\cos ^2 x = \hfill \\ = 1 - 2\left( {\sin x\cos x} \right)^2 . \hfill \\ \end{gathered}$

$\int {\frac{{\sin x\cos x}}{{\sin ^4 x + \cos ^4 x}}dx} = \int {\frac{{\sin x\cos x}}{{1 - 2\left( {\sin x\cos x} \right)^2 }}dx = }$

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

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

That's what I was thinking. Thanks for putting in the details.

DeMath · #9 $Old$ January 3rd, 2009, 03:43 PM

2. $\begin{gathered}\int {\frac{{\arctan e^x }}{{\cosh x}}dx} = 2\int {\frac{{\arctan e^x }}{{e^x + e^{ - x} }}dx} = 2\int {\frac{{e^x }}{{e^{2x} + 1}}\arctan e^x dx} = \hfill \\= 2\int {\arctan e^x d\left( {\arctan e^x } \right)} = \arctan ^2 e^x + C. \hfill \\ \end{gathered}$

DeMath · #10 $Old$ January 3rd, 2009, 04:10 PM

3.
$\int {\frac{{\cos xdx}}{{\sqrt {1 + \sin x + \cos ^2 x} }}} = \int {\frac{{\cos xdx}}{{\sqrt {2 + \sin x - \sin ^2 x} }}} = \left\{ \begin{gathered}\sin x = t, \hfill \\\cos xdx = dt \hfill \\ \end{gathered} \right\} =$

$\int {\frac{{dt}}{{\sqrt {2 + t - t^2 } }}} = 2\int {\frac{{dt}} {{\sqrt {8 + 4t - 4t^2 } }}} = 2\int {\frac{{dt}}{{\sqrt {9 - \left( {4t^2 - 4t + 1} \right)} }}} =$

$= 2\int {\frac{{dt}}{{\sqrt {9 - \left( {2t - 1} \right)^2 } }}} = \frac{2}{3}\int {\frac{{dt}}{{\sqrt {1 - \left( {\frac{{2t - 1}}{3}} \right)^2 } }}} = \int {\frac{{d\left( {\frac{{2t - 1}}{3}} \right)}}{{\sqrt {1 - \left( {\frac{{2t - 1}}{3}} \right)^2 } }}} =$

$= \arcsin \left( {\frac{{2t - 1}}{3}} \right) + C = \arcsin \left( {\frac{{2\sin x - 1}}{3}} \right) + C.$

DeMath · #11 $Old$ January 3rd, 2009, 04:39 PM

Quote:

Originally Posted by Ment $View Post$

Please, help me solution these integrals:

$1.{\text{ }}\int {\frac{{\sin x\cos x}} {{\sin ^4 x + \cos ^4 x}}dx} .$

$2.{\text{ }}\int {\frac{{\arctan e^x }}{{\cosh x}}} dx.$

$3.{\text{ }}\int {\frac{{\cos x}}{{\sqrt {1 + \sin x + \cos ^2 x} }}dx} .$

$4.{\text{ }}\int {\frac{{dx}}{{a + b\cos x}}} {\text{ if }}1){\text{ }}a > 0{\text{ }} \wedge {\text{ }}b > 0;{\text{ }}2){\text{ }}a > 0{\text{ }} \wedge {\text{ }}b < 0.$

4. $\int {\frac{{dx}}{{a + b\cos x}}} .$
We will consider the first case $a > 0$ and $b > 0$ and $a > b$ .

$\begin{gathered}a + b\cos x = a\left( {\sin ^2 \frac{x}{2} + \cos ^2 \frac{x}{2}} \right) + b\cos ^2 \frac{x}{2} - b\sin ^2 \frac{x}{2} = \hfill \\= \left( {a - b} \right)\sin ^2 \frac{x}{2} + \left( {a + b} \right)\cos ^2 \frac{x}{2}. \hfill \\ \end{gathered}$

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

$= \frac{1}{{a + b}}\int {\frac{{\frac{1}{{\cos ^2 \frac{x}{2}}}}}{{1 + \left( {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{x}{2}} \right)^2 }}dx} = \frac{2}{{\sqrt {a^2 - b^2 } }}\int {\frac{{d\left( {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{x}{2}} \right)}}{{1 + \left( {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{x}{2}} \right)^2 }}} =$

$= \frac{2}{{\sqrt {a^2 - b^2 } }}\arctan \left( {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{x}{2}} \right) + C.$

Danny · #12 $Old$ January 3rd, 2009, 04:59 PM

Quote:

Originally Posted by DeMath $View Post$

3.
$\int {\frac{{\cos xdx}}{{\sqrt {1 + \sin x + \cos ^2 x} }}} = \int {\frac{{\cos xdx}}{{\sqrt {2 + \sin x - \sin ^2 x} }}} = \left\{ \begin{gathered}\sin x = t, \hfill \\\cos xdx = dt \hfill \\ \end{gathered} \right\} =$

$\int {\frac{{dt}}{{\sqrt {2 + t - t^2 } }}} = 2\int {\frac{{dt}}{{\sqrt {8 + 4t - 4t^2 } }}} = 2\int {\frac{{dt}}{{\sqrt {9 - \left( {4t^2 - 4t + 1} \right)} }}} =$

$= 2\int {\frac{{dt}}{{\sqrt {9 + \left( {2t - 1} \right)^2 } }}} = \frac{2}{3}\int {\frac{{dt}}{{\sqrt {1 + \left( {\frac{{2t - 1}}{3}} \right)^2 } }}} = \int {\frac{{d\left( {\frac{{2t - 1}}{3}} \right)}}{{\sqrt {1 + \left( {\frac{{2t - 1}}{3}} \right)^2 } }}} =$

$= \arcsin \left( {\frac{{2t - 1}}{3}} \right) + C = \arcsin \left( {\frac{{2\sin x - 1}}{3}} \right) + C.$

Sign problem in some of your integrals (answers right though)

DeMath · #13 $Old$ January 3rd, 2009, 05:20 PM

Quote:

Originally Posted by danny arrigo $View Post$

Sign problem in some of your integrals (answers right though)

Thank you, I understood

. I made the mistakes in radicals here

$\color{red}2\int {\frac{{dt}}{{\sqrt {9 + \left( {2t - 1} \right)^2 } }}} = \frac{2}{3}\int {\frac{{dt}}{{\sqrt {1 + \left( {\frac{{2t - 1}}{3}} \right)^2 } }}} = \int {\frac{{d\left( {\frac{{2t - 1}}{3}} \right)}}{{\sqrt {1 + \left( {\frac{{2t - 1}}{3}} \right)^2 } }}}$

Minus is necessary

$2\int {\frac{{dt}}{{\sqrt {9 - \left( {2t - 1} \right)^2 } }}} = \frac{2}{3}\int {\frac{{dt}}{{\sqrt {1 - \left( {\frac{{2t - 1}}{3}} \right)^2 } }}} = \int {\frac{{d\left( {\frac{{2t - 1}}{3}} \right)}}{{\sqrt {1 - \left( {\frac{{2t - 1}}{3}} \right)^2 } }}}$