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Evaluate the integral. \int_{2\sqrt 3}^{6} \frac{1}{t^3 \sqrt{t^2-9}} dt

Question:

Evaluate the integral.

{eq}\int_{2\sqrt 3}^{6} \frac{1}{t^3 \sqrt{t^2-9}} {/eq} dt

Definite Integrals:

Here the definite integral is having the square root terms in the denominator, so we will first simplify the integrand expression before, we can apply the trigonometric substitution method. to solve the integrals.

Answer and Explanation:


Given integral is :

{eq}\int_{2\sqrt3}^{6} \frac{1}{t^3 \sqrt{t^2-9}} dt\\ \mathrm{Compute\:the\:indefinite\:integral}:\quad \int \frac{1}{t^3\sqrt{t^2-9}}dt\\ \mathrm{Apply\:Trig\:Substitution:}\:t=3\sec \left(u\right)\\ => \int \frac{1}{27\sec ^2\left(u\right)}du\\ =\frac{1}{27}\cdot \int \:\cos ^2\left(u\right)du\\ =\frac{1}{27}\cdot \int \:\frac{1+\cos \left(2u\right)}{2}du\\ =\frac{1}{27}\cdot \frac{1}{2}\left(u+\frac{1}{2}\sin \left(2u\right)\right)+c\\ \mathrm{Substitute\:back}\:u=\sec^{-1} \left(\frac{1}{3}t\right)\\ =\frac{1}{54}\left(\sec^{-1} \left(\frac{1}{3}t\right)+\frac{1}{2}\sin \left(2\sec^{-1} \left(\frac{1}{3}t\right)\right)\right)+C\\ \mathrm{Compute\:the\:boundaries}:\quad \int _{2\sqrt{3}}^6\frac{1}{t^3\sqrt{t^2-9}}dt\\ =\left [ \frac{1}{54}\left(\sec^{-1} \left(\frac{1}{3}t\right)+\frac{1}{2}\sin \left(2\sec^{-1} \left(\frac{1}{3}t\right)\right)\right) \right ]_{2\sqrt3}^{6}\\ =\frac{4\pi +3\sqrt{3}}{648}-\frac{2\pi +3\sqrt{3}}{648}\\ =\frac{\pi }{324} {/eq}


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