2.9 Calculus of Hyperbolic Functions

[STARTS AT 7:58]

INSTRUCTOR: OK, number 61. u substitution would be very handy here because we have an inside function, sine cubed. And the derivative of that is cosine. So we can make u sine hx. du is then cosine h x dx. So this becomes the integral of u cubed times du, which is u to the fourth over 4 plus c. And since we know what u is, this will be sine h to the fourth of x over 4 plus c. 

OK, next, last integral of this form we're going to have-- kind of flying through this-- is a hyperbolic secant. All right, so first, we should notice we have an inside function, first of all. That is 3x, which would make du 3 dx, or 1/3 du equals dx. 

So writing this in terms of u, we get the integral of secant hyperbolic secant squared u du. And I'll bring my factor of 1/3 there. Right? Now, I'm just fine with having a secant squared because the integral of secant squared is tangent. So this is 1/3 hyperbolic tangent of u. And I'm going to go ahead and substitute it back in because u is 3x plus c. 

We always need to make sure we adjust for that inside function we've got.

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