Question

$\mathbf{\int }\sqrt{\mathbf{1}\mathbf{+}\mathbf{"}\mathbf{"}\mathbf{a}{\mathbf{"}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{x}}{\mathbf{2}}\sqrt{\mathbf{1}\mathbf{+}\mathbf{"}\mathbf{"}\mathbf{a}{\mathbf{"}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{a}}\mathit{s}\mathit{i}\mathit{n}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

Short answer

The required solution is$\int \sqrt{1+x^2{a}^2)}dx=\frac{1}{2a}\left[\sin h^{-1}\left(ax\right)+1+x^2{a}^{2}xa\right]+c$

Step-by-step solution

Definition of definite integral

A function that practices the antiderivative of another function is called an indefinite integral. It can be represented graphically as an integral symbol, a function, and finally a dx. The indefinite integral is a more straightforward way of expressing the antiderivative.

Using the substitution

Make use of the substitute$x=\left(\frac{1}{2}\right)\sin hu$

role="math" localid="1658836295640" $$\begin{gathered} \int \sqrt{1+x^2{a}^2}dx=\int \sqrt{1+\sin h^{2}u}\frac{1}{a}\cos hudu \\ =\frac{1}{a}\int \cos h^{2}udu \\ =\frac{1}{a}\int \frac{1}{2}\left(\cos h\left(2u\right)+1\right)du \\ =\frac{1}{2a}\left[u+\int \cos h\left(2u\right)du\right] \\ =\frac{1}{2a}\left[u+\frac{1}{2}\sin h\left(2u\right)\right]+C \\ =\frac{1}{2a}\left[u+\cos hu\sin hu\right]+C \\ =\frac{1}{2}\left[\sin h^{-1}\left(ax\right)+\sqrt{1+\sin h^{2}uxa}\right]+C \\ =\frac{1}{2}\left[\sin h^{-1}\left(ax\right)+\sqrt{1+x^2{a}^{2}xa}\right]+C \end{gathered}$$

Now, Apply role="math" localid="1658836375727" $cos{h}^2"x-sin{h}^{2}x=1andcos{h}^{2}x=(1+cosh(2x\left)\right)/2.$

Therefore, the answer to the question is

$\int \sqrt{1+x^2{a}^2}dx=\frac{1}{2a}\left[\sin h^{-1}\left(ax\right)+\sqrt{1+x^2{a}^2}xa\right]+C$