Trig substitution integral letting x=tan(theta): integrate 1/(1+x^2)^(3/2).
969 بار بازدید -
2 سال پیش
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We integrate 1/(1+x^2)^(3/2) by using
We integrate 1/(1+x^2)^(3/2) by using trigonometric substitution. In this case, we make progress by letting x=tan(theta) and dx=sec^2(theta)d(theta). We plan to exploit the Pythagorean identity 1+tan^2(theta)=sec^2(theta) to simplify the denominator.
We make the trig substitutions for x and dx and transform the integral to theta space, where the integrand simplifies to cos(theta). The integral is computed as sin(theta)+C, but we aren't done yet! We have to transform the antiderivative in terms of x, and to get this done, we revisit the original substitution x=tan(theta) and solve for theta=arctan(x), or the angle whose tangent is x.
Now the result of the integral looks like sin(arctan(x))+C, but we still aren't done! We can always evaluate a trig function of an inverse trig function into an algebraic form, and we do this by visualizing a right triangle with the angle theta=angle whose tangent is x in it. This allows us to label the opposite side as x, the adjacent side as 1 and compute the hypotenuse as sqrt(1+x^2). Now we can finally compute the sin(tan^(-1)x) as x/sqrt(1+x^2), and we're done with our trig substitution integral letting x=tan(theta).
We make the trig substitutions for x and dx and transform the integral to theta space, where the integrand simplifies to cos(theta). The integral is computed as sin(theta)+C, but we aren't done yet! We have to transform the antiderivative in terms of x, and to get this done, we revisit the original substitution x=tan(theta) and solve for theta=arctan(x), or the angle whose tangent is x.
Now the result of the integral looks like sin(arctan(x))+C, but we still aren't done! We can always evaluate a trig function of an inverse trig function into an algebraic form, and we do this by visualizing a right triangle with the angle theta=angle whose tangent is x in it. This allows us to label the opposite side as x, the adjacent side as 1 and compute the hypotenuse as sqrt(1+x^2). Now we can finally compute the sin(tan^(-1)x) as x/sqrt(1+x^2), and we're done with our trig substitution integral letting x=tan(theta).
2 سال پیش
در تاریخ 1401/09/08 منتشر شده
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969
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