(d) ( int1^3 fracd xsqrtx(1+x) ) (d) ( intsqrt3^3...

Transcript

00:01 You want to evaluate this integral.

00:02 I'm going to make a substitution u is equal to the square root of x, or x equals u squared.

00:13 So dx is 2 u, d u, so this will be the integral from 1 to the square root of 3, and then our d x is 2 u, and then the denominator we've got u times 1 plus 3 plus u squared.

00:43 So this comes out to be two integral one to the square of three of d u over one plus u squared.

00:58 So that's equal to we've got a factor of two out in front.

01:03 Of course that gives us an inverse tangent of you from one to the square of three.

01:14 So we get this.

01:16 So the inverse tangent of the square of three is pi over six and the inverse tangent of one is pi over four comes out to be minus pi over six okay so this one will probably also involve an inverse tangent but i want to do a substitution to make sure that we get all the factors correct so we got x is going to be tangent theta so d x is ccan squared theta theta the dena of three right then the denominator is going to be nine times ccquare theta all right so this will be 12 over three and then we'll integrate from three times the inverse tangent of live three to three times the inverse tangent of three then we got d -theta.

03:39 So that's equal to four times, actually there's another is three, comes from there.

03:50 We've got the inverse tangent of three minus the inverse tangent of a third.

04:01 So we just leave it in that form.

04:03 We know that the inverse tangent of a third, we just figured it out, is kind of nice, but the inverse tangent of 3 is a mess.

04:13 So this comes out to be 12 out in front...

Question

(d) ( int_{1}^{3} frac{d x}{sqrt{x}(1+x)} ) (d) ( int_{sqrt{3}}^{3} frac{12}{9+x^{2}} d x ) [ int_{-frac{pi}{2}}^{frac{pi}{2}} frac{3 cos x}{1+sin ^{2} x} d x ]

Question

(d) ( int_{1}^{3} frac{d x}{sqrt{x}(1+x)} ) (d) ( int_{sqrt{3}}^{3} frac{12}{9+x^{2}} d x ) [ int_{-frac{pi}{2}}^{frac{pi}{2}} frac{3 cos x}{1+sin ^{2} x} d x ]

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