Here, we will discuss the second set of standard integrals which can be evaluated by using integration by parts. This is a formula given by \[\int (uv)dx=u\int vdx-\int\left(\frac{du}{dx}\int vdx\right)dx\] With the application of this formula, we can evaluate the following standard integrals.
$\int e^{ax}\cos bxdx=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C$
\[\text{Let }I=\int e^{ax}\cos bxdx\] \[=e^{ax}\int\cos bxdx-\int\left(\frac{d}{dx}(e^{ax})\int\cos bxdx\right)dx\] \[=\frac{e^{ax}\sin bx}{b}-\frac{a}{b}\int e^{ax}\sin bxdx\] \[=\frac{e^{ax}\sin bx}{b}-\frac{a}{b}\left[e^{ax}\int\sin bxdx-\int\left(\frac{d}{dx}(e^{ax})\int\sin bxdx\right)dx\right]\] \[=\frac{e^{ax}\sin bx}{b}+\frac{a}{b^2}e^{ax}\cos bx-\frac{a^2}{b^2}\int e^{ax}\cos bxdx\] \[\text{or, I}=\frac{e^{ax}\sin bx}{b}+\frac{a}{b^2}e^{ax}\cos bx-\frac{a^2}{b^2}I\] \[\text{or, }\frac{a^2+b^2}{b^2}I=e^{ax}\frac{(b\sin bx+a\cos bx)}{b^2}\] \[\text{or, }I=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C\]
$\int e^{ax}\sin bx dx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C$
\[\text{Let }I=e^{ax}\sin bxdx\] \[=e^{ax}\int \sin bx-\int\left(\frac{d}{dx}(e^{ax})\int\sin bxdx\right)dx\] \[=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}\int e^{ax}\cos bxdx\] \[=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}\left[e^{ax}\int\cos bxdx-\int\left(\frac{d}{dx}(e^{ax})\int\cos bxdx\right)dx\right]\] \[=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b^2}e^{ax}\sin bx-\frac{a^2}{b^2}\int e^{ax}\sin bxdx\] \[\text{or, }I=\frac{-e^{ax}\cos bx}{b}+\frac{a}{b^2}e^{ax}\sin bx-\frac{a^2}{b^2}I\] \[\text{or, }I=\frac{a^2+b^2}{b^2}I=\frac{e^{ax}(-b\cos bx+a\sin bx)}{b^2}\] \[\text{or, }I=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C\]
$\int\sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log(x+\sqrt{x^2+a^2})+C$
\[\text{Let }I=\int\sqrt{x^2+a^2}dx\] \[=\sqrt{x^2+a^2}\int 1dx-\int\left(\frac{d}{dx}\left(\sqrt{x^2+a^2}\right)\int 1dx\right)dx\] \[=x\sqrt{x^2+a^2}-\int\frac{2x}{2\sqrt{x^2+a^2}}xdx\] \[=x\sqrt{x^2+a^2}-\int\frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}dx\] \[=x\sqrt{x^2+a^2}-\int\sqrt{x^2+a^2}dx+a^2\int\frac{dx}{x^2+a^2}\] \[\text{or, }I=x\sqrt{x^2+a^2}-I+a^2\log(x+\sqrt{x^2+a^2})\] \[\text{or, }2I=x\sqrt{x^2+a^2}+a^2\log(x+\sqrt{x^2+a^2})\] \[\text{or, }I=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log(x+\sqrt{x^2+a^2})+C\]
$\int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log(x+\sqrt{x^2-a^2})+C$
\[\text{Let }I=\int\sqrt{x^2-a^2}dx\] \[=\sqrt{x^2-a^2}\int 1dx-\int\left(\frac{d}{dx}\left(\sqrt{x^2-a^2}\right)\int 1dx\right)dx\] \[=x\sqrt{x^2-a^2}-\int\frac{2x}{2\sqrt{x^2-a^2}}xdx\] \[=x\sqrt{x^2-a^2}-\int\frac{(x^2-a^2)+a^2}{\sqrt{x^2-a^2}}dx\] \[=x\sqrt{x^2-a^2}-\int\sqrt{x^2-a^2}dx-a^2\int\frac{dx}{\sqrt{x^2-a^2}}\] \[\text{or, }I=x\sqrt{x^2-a^2}-I-a^2\log(x+\sqrt{x^2-a^2})\] \[\text{or, }2I=x\sqrt{x^2-a^2}-a^2\log(x+\sqrt{x^2-a^2})\] \[\text{or, }I=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log(x+\sqrt{x^2-a^2})+C\]
$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$
\[\text{Let }I=\int\sqrt{a^2-x^2}dx\] \[=\sqrt{a^2-x^2}\int 1dx-\int\left(\frac{d}{dx}\left(\sqrt{a^2-x^2}\right)\int 1dx\right)dx\] \[=x\sqrt{a^2-x^2}-\int\frac{-2x}{2\sqrt{a^2-x^2}}xdx\] \[=x\sqrt{a^2-x^2}-\int\frac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx\] \[=x\sqrt{a^2-x^2}-\int\sqrt{a^2-x^2}dx+a^2\int\frac{dx}{\sqrt{a^2-x^2}}\] \[\text{or, }I=x\sqrt{a^2-x^2}-I+a^2\sin^{-1}\frac{x}{a}\] \[\text{or, }2I=x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\] \[\text{or, }I=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\]
Some Integrals Reducible to Standard Forms
The integrals of the form \[\int\sqrt{ax^2+bx+c}\text{ }dx\] can now be easily integrated by converting them into one of the above standard integrals. The other form of the integral which can be reduced to the sum of two integrable forms is \[\int(mx+e)\sqrt{ax^2+bx+c}\text{ }dx\text{ __(1)}\]
This form can be written as the sum of two integrals \[p\int(2ax+b)\sqrt{ax^2+bx+c}\text{ }dx+q\int\sqrt{ax^2+bx+c}\text{ }dx\] in which the integrand of the first integral is the product $2ax+b$ and $\sqrt{ax^2+bx+c}$ and $2ax+b$ is the differential coefficient of $ax^2+bx+c$ and the constants $p$ and $q$ are to be adjusted so as to make the sum equal to the given integral $\text{(1)}$. The values of $p$ and $q$ can be obtained from $2ap=m$ and $bp+q=e$.
$\int\sqrt{2ax-x^2}\text{ }dx$
\[\int\sqrt{2ax-x^2}\text{ }dx=\int\sqrt{a^2-a^2+2ax-x^2}\text{ }dx\] \[=\int\sqrt{a^2-(x-a)^2}\text{ }dx\] \[=\frac{x-a}{2}\sqrt{2ax-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x-a}{a}\right)+C\]
$\int\sqrt{4x^2-4x+5}\text{ }dx$
\[\text{Let }I=\int\sqrt{4x^2-4x+5}\text{ }dx\] \[=\int\sqrt{4x^2-4x+1+4}\text{ }dx\] \[=\int\sqrt{(2x-1)^2+2^2}\text{ }dx\] Put $2x=y$, then, $dx=\frac{dy}{2}$. \[\therefore I=\frac{1}{2}\int\sqrt{(y-1)^2+2^2}\text{ }dy\] \[=\frac{y-1}{4}\sqrt{4x^2-4x+5}+\log(y-1+\sqrt{4x^2-4x+5})+C\] \[=\frac{2x-1}{4}\sqrt{4x^2-4x+5}+\log(2x-1+\sqrt{4x^2-4x+5})+C\]
$\int(2x+1)\sqrt{4x^2+20x+21}\text{ }dx$
\[\text{Let }I=\int(2x-1)\sqrt{4x^2+20x+21}\text{ }dx\] \[=\frac{1}{4}\int(8x+4)\sqrt{4x^2+20x+21}\text{ }dx\] \[=\frac{1}{4}\int(8x+20-16)\sqrt{4x^2+20x+21}\text{ }dx\] \[=\frac{1}{4}\int(8x+20)\sqrt{4x^2+20x+21}\text{ }dx-\frac{16}{4}\int\sqrt{4x^2+20x+21}\text{ }dx\] \[=\frac{1}{4}I_1-4I_2\] where, \[\frac{1}{4}I_1=\int(8x+20)\sqrt{4x^2+20x+21}\text{ }dx\]
Put $4x^2+20x+21=y$, then, $(8x+20)dx=dy$. \[\therefore \frac{1}{4}I_1=\frac{1}{4}\int\sqrt{y}\text{ }dy=\frac{1}{4}.\frac{2}{3}y^{\frac{3}{2}}+C_1\] \[=\frac{1}{16}(4x^2+20x+21)^{\frac{3}{2}}+C_1\] and, \[4I_1=4\int\sqrt{4x^2+20x+21}\text{ }dx\] \[=4\int\sqrt{4x^2+20x+25-4}\text{ }dx\] \[=4\int\sqrt{(2x+5)^2-4}\text{ }dx\]
Put $2x=z$, then, $dx=\frac{dz}{2}$. \[\therefore 4I_2=\frac{4}{2}\int\sqrt{(x+5)^2-4}\text{ }dz\] \[=2\int\sqrt{(z+5)^2-2^2}\text{ }dz\] \[=2\left[\frac{z+5}{2}\sqrt{4x^2+20x+21}-\frac{4}{2}\log\sqrt{4x^2+20x+21}\right]+C_2\] \[=(2x+5)\sqrt{4x^2+20x+21}-4\log\sqrt{4x^2+20x+21}+C_2\]
Putting the values of $\frac{1}{4}I_1$ and $4I_2$, \[\therefore I=\frac{1}{16}(4x^2+20x+21)^{\frac{3}{2}}-(2x+5)\sqrt{4x^2+20x+21}\] \[+4\log\sqrt{4x^2+20x+21}+C\]
More Standard Integrals