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W3bbo
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[SOLVED] Some antiderivatives
I've got a few antiderivatives to find, I've found most of them and they check out fine with my CAS, but three of them I'm having difficulties with.
The first:
[tex]I = \int {{{\sec ^2 \left( x \right)} \over {\left( {1 + \tan \left( x \right)} \right)^3 }}} dx[/tex]
Using Integration by Substitution:
[tex]
\displaylines{
I = \int {{{\sec ^2 \left( x \right)} \over {\left( {1 + \tan \left( x \right)} \right)^3 }}} dx \cr
u = 1 + \tan \left( x \right) \cr
{{du} \over {dx}} = \sec ^2 \left( x \right) \cr
du = \sec ^2 \left( x \right)dx \cr
I = \int {{{\sec ^2 \left( x \right)} \over {u^3 }}} dx \cr
= \int {u^{ - 3} } du \cr
= - {\textstyle{1 \over 2}}u^{ - 2} \cr
= - {\textstyle{1 \over 2}}\left( {1 + \tan \left( x \right)} \right)^{ - 2} + C \cr
= - {1 \over {2\left( {1 + \tan \left( x \right)} \right)^2 }} + C \cr}
[/tex]
Yet this does not coincide with the answer my CAS gets:
[tex]{{ - {\mathop{\rm Cos}\nolimits} \left( {2x} \right) + {\mathop{\rm Sin}\nolimits} \left( {2x} \right)} \over {4 + 8{\mathop{\rm Cos}\nolimits} \left( x \right){\mathop{\rm Sin}\nolimits} \left( x \right)}}[/tex]
Finally, for the other two problems I can't think of an approach. I've tried Integration by Parts on both of them, to no avail. I've also tried Substitution, but I get stuck quickly.
FWIW, they're:
[tex]I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx[/tex]
And
[tex]I = \int {{\mathop{\rm Cos}\nolimits} ^2 \left( {2x} \right)} dx[/tex]
Thanks!
I've got a few antiderivatives to find, I've found most of them and they check out fine with my CAS, but three of them I'm having difficulties with.
The first:
Homework Statement
[tex]I = \int {{{\sec ^2 \left( x \right)} \over {\left( {1 + \tan \left( x \right)} \right)^3 }}} dx[/tex]
The Attempt at a Solution
Using Integration by Substitution:
[tex]
\displaylines{
I = \int {{{\sec ^2 \left( x \right)} \over {\left( {1 + \tan \left( x \right)} \right)^3 }}} dx \cr
u = 1 + \tan \left( x \right) \cr
{{du} \over {dx}} = \sec ^2 \left( x \right) \cr
du = \sec ^2 \left( x \right)dx \cr
I = \int {{{\sec ^2 \left( x \right)} \over {u^3 }}} dx \cr
= \int {u^{ - 3} } du \cr
= - {\textstyle{1 \over 2}}u^{ - 2} \cr
= - {\textstyle{1 \over 2}}\left( {1 + \tan \left( x \right)} \right)^{ - 2} + C \cr
= - {1 \over {2\left( {1 + \tan \left( x \right)} \right)^2 }} + C \cr}
[/tex]
Yet this does not coincide with the answer my CAS gets:
[tex]{{ - {\mathop{\rm Cos}\nolimits} \left( {2x} \right) + {\mathop{\rm Sin}\nolimits} \left( {2x} \right)} \over {4 + 8{\mathop{\rm Cos}\nolimits} \left( x \right){\mathop{\rm Sin}\nolimits} \left( x \right)}}[/tex]
Finally, for the other two problems I can't think of an approach. I've tried Integration by Parts on both of them, to no avail. I've also tried Substitution, but I get stuck quickly.
FWIW, they're:
[tex]I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx[/tex]
And
[tex]I = \int {{\mathop{\rm Cos}\nolimits} ^2 \left( {2x} \right)} dx[/tex]
Thanks!
Last edited: