Asked by karan
How do I integrate tanx (1+sec^4 x)^3/2 dx
My daughter is doing surface area problems, and all the examples assume she knows how to finish it off once we get to the integration....
My daughter is doing surface area problems, and all the examples assume she knows how to finish it off once we get to the integration....
Answers
Answered by
Helper
All the online integration calculators I tried said that this cannot be integrated.
Recheck your problem.
Recheck your problem.
Answered by
MathMate :)
Try substitution:
u=cos(x)
du=-sin(x)dx
Then
∫tanx (1+sec^4 x)^3/2 dx
=∫(1+sec^4 x)^3/2 sin(x) dx / cos(x)
=-∫[(1+(1/u)^4)^3/2 /u] du
But the integration is still ugly!
Rechecking the problem is the first thing to do.
u=cos(x)
du=-sin(x)dx
Then
∫tanx (1+sec^4 x)^3/2 dx
=∫(1+sec^4 x)^3/2 sin(x) dx / cos(x)
=-∫[(1+(1/u)^4)^3/2 /u] du
But the integration is still ugly!
Rechecking the problem is the first thing to do.
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