91Ó°ÊÓ

#tag_title#Step 4: Evaluate the individual integrals and combine#tag_content# For the first integral: $$\int \cos^2{t} d t = \int \frac{1 + \cos{2t}}{2} d t = \frac{1}{2} \int (1 + \cos{2t}) d t = \frac{1}{2} (t + \frac{1}{2}\sin{2t}) + C_1.$$ For the second integral: $$\int \frac{\cosh^2{u}}{\cosh^2{u} \cosh{u}}d u - \int \frac{1}{\cosh^2{u} \cosh{u}} d u = \int \frac{1}{\cosh{u}} d u - \int \frac{1}{\cosh^3{u}} d u.$$ The first part can be solved directly: $$\int \frac{1}{\cosh{u}} d u = \int \frac{\cosh{u}}{\cosh^2{u}} d u = \int \frac{\cosh{u}}{\cosh^2{u} - \sinh^2{u}} d u = \int \operatorname{sech}u \,d u.$$ This integral can be solved, and its result is: $$\int \operatorname{sech}u \,d u = \arctan(\sinh u) + C_2.$$ Now, let's rewrite the result using the substitution we made at the beginning: $$\int \operatorname{sech}u \,d u = \arctan(\sinh u) + C_2 = \arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C_2.$$ Now, we can combine the results of the first and second integral: $$\int \frac{1}{\cosh^{3}{u}}d u = \frac{1}{2} (t + \frac{1}{2}\sin{2t}) + C_1 - \left(\arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C_2\right).$$ Finally, we can change the variable \(t\) back to \(u\), using \(\cosh{u} = \sec{t}\) so that \(\arccos{(\cosh{u})} = t\), to get the final answer: $$\int \frac{1}{x^{3} \sqrt{x^{2}-100}}d x = \frac{1}{2} \left(\arccos{(\cosh{u})} + \frac{1}{2}\sin{2 \arccos{(\cosh{u})}}\right) - \arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C_3.$$ So the final answer is: $$\int \frac{1}{x^{3} \sqrt{x^{2}-100}}d x = \frac{1}{2} \left(\arccos{(\cosh{u})} + \frac{1}{2}\sin{2 \arccos{(\cosh{u})}}\right) - \arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C.$$