Find the integral using an appropriate trigonometric...

Find the integral using an appropriate trigonometric substitution.\\ $\int \frac{x^3}{\sqrt{x^2+4}} dx$\\ $\frac{1}{3}(x^2+4)^{3/2}\sqrt{x^2+4} + C$\ $\frac{1}{3}(x^2-8)\sqrt{x^2+4} + C$\ $\frac{1}{3}(x^2+8)\sqrt{x^2+4} + C$\ $\frac{1}{3}(x^2-4)^{3/2}\sqrt{x^2+4} + C$

Transcript

00:01 In this question we have given four integrals and we have to find the appropriate isometric substitution in the form of x is equals to f of t so that using this substitution we can find the integral so the first integral a is given us integration of 4x square minus 7 raised to the power 3 divided by 2 d x so here here we have to write the appropriate substitution in the form of x is equal to f of t so the substitution is x is equal to under root of 7 multiplied with sec t divided by 2 now we have given a second integral s b that is x squared divided by under root of 8 x square plus 6 d x now, in order to simplify this integral, we will use a substitution.

01:06 X is equal to root 3 multiplied with 10t divided by.

01:15 Now, the third integral is given as t, that is integration of x multiplied with under root of 8x square plus 80x plus 194 d.

01:33 Here we are not introducing the x substitution as a trigonometric function in the beginning.

01:40 First we have to write this expiration in more simplified way.

01:45 So this can be written as integration of 8x plus 40 multiplied it with under root of 4x square plus 40x plus 97 whole divided by 8 minus integration of 5 under root of 4x square plus 40x plus 97 d x now in order to integrate the second term we will substitute first we will write it as integration under root of 2x plus 10 whole square minus 3 d 8 after this we can rewrite it as under root of 4 times x plus 5 whole square minus 3 d x and then substitute u is equal to x plus 5 after substituting this we will get it as integration of under root 4 u square minus 3 d x now substitute u is equals to arc sec 2 u divided by under root of 3 so this is the substitution of trigonometric function in this integration...

Question

Find the integral using an appropriate trigonometric substitution.\\ $\int \frac{x^3}{\sqrt{x^2+4}} dx$\\ $\frac{1}{3}(x^2+4)^{3/2}\sqrt{x^2+4} + C$\ $\frac{1}{3}(x^2-8)\sqrt{x^2+4} + C$\ $\frac{1}{3}(x^2+8)\sqrt{x^2+4} + C$\ $\frac{1}{3}(x^2-4)^{3/2}\sqrt{x^2+4} + C$

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