Question

The antiderivative of

Answer 1

Let $A=\int \frac{1}{(a^2+b^2)-(a^2+b^2)\cos x}dx$
Substituting $t=\cos x\Rightarrow dt=-\sin xdx$ we get $A=-\frac{1}{(a^2+b^2)}\int \frac{1}{(t-1)\sqrt{1-t^2}}dt$
Again substituting $t=\sin u\Rightarrow dt=\cos udu$, we get $A=-\frac{1}{(a^2+b^2)}\int \frac{1}{\sin u-1}du$
Now substituting $s=\tan \frac{u}{2}\Rightarrow ds=\frac{1}{2}{\sec }^2\frac{u}{2}du$, we get
$A=\frac{2}{(a^2+b^2)}\int \frac{1}{{(s-1)}^2}=-\frac{2}{(a^2+b^2)(s-1)}+c=-\frac{2}{(a^2+b^2)(\tan \frac{u}{2}-1)}+c=-\frac{2}{(a^2+b^2)(\frac{t}{\sqrt{1-t^2}+1}-1)}+c=-\frac{2}{(a^2+b^2)(\frac{\cos x}{\sqrt{1-{\cos }^{2}x}+1}-1)}+c=-\frac{2}{(a^2+b^2)(\frac{\cos x}{\sqrt{1-{\cos }^{2}x}+1}-1)}+c=-\frac{\cot \frac{x}{2}}{(a^2+b^2)}$
Let $B=\int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}dx$
Multiplying and dividing by ${\sec }^{2}x$ we get $B=\int \frac{{\sec }^{2}x}{a^2{\tan }^{2}x+b^2}dx$
Substituting $t=\tan x\Rightarrow dt={\sec }^{2}xdx$, we get
$B=\int \frac{1}{a^2{t}^2+b^2}dt=\frac{1}{b^2}\int \frac{1}{\frac{a^2{t}^2}{b^2}+1}dt=\frac{{\tan }^{-1}\frac{at}{b}}{ab}+c=\frac{1}{ab}{\tan }^{-1}\frac{a\tan x}{b}+c$
Let $C=\int \frac{1}{a\cos x+b\sin x}dx$
Substituting $t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx$, we get
$C=2\int \frac{1}{-a{t}^2+a+2bt}=2\int \frac{1}{\frac{b^2}{a}-{(\sqrt{a}t-\frac{b}{\sqrt{a}})}^2+a}dt$
Again substituting $u=\sqrt{a}t-\frac{b}{\sqrt{a}}\Rightarrow du=\sqrt{a}dt$, we get
$C=\frac{2}{\sqrt{a}}\int \frac{1}{\frac{b^2}{a}+a-u^2}du=\frac{2}{\sqrt{a}(\frac{b^2}{a}+a)}\int \frac{1}{1-\frac{a{u}^2}{a^2+b^2}}du=\frac{2}{\sqrt{a^2+b^2}}\log \frac{1+\frac{\sqrt{a}u}{\sqrt{a^2+b^2}}}{1-\frac{\sqrt{a}u}{\sqrt{a^2+b^2}}}+c=\frac{2}{\sqrt{a^2+b^2}}\log \frac{\sqrt{a^2+b^2}+at-b}{\sqrt{a^2+b^2}-at+b}+c=\frac{2}{\sqrt{a^2+b^2}}\log \frac{\sqrt{a^2+b^2}+a\tan \frac{x}{2}-b}{\sqrt{a^2+b^2}-a\tan \frac{x}{2}+b}+c$
Let $D=\int \frac{1}{a^2-b^2{\cos }^{2}x}dx$
Multiplying and dividing by ${\sec }^{2}x$, we get $D=\int \frac{{\sec }^{2}x}{a^2{\sec }^{2}x-b^2}dx$
Now substituting $t=\tan x\Rightarrow dt={\sec }^{2}xdx$, we get
$D=\int \frac{1}{a^2(t^2+1)-b^2}dt=\frac{1}{a^2-b^2}\int \frac{1}{\frac{a^2{t}^2}{a^2-b^2}-1}dt=\frac{1}{a\sqrt{a^2-b^2}}{\tan }^{-1}\left(\frac{at}{\sqrt{a^2-b^2}}\right)+c=\frac{1}{a\sqrt{a^2-b^2}}{\tan }^{-1}\left(\frac{a\tan x}{\sqrt{a^2-b^2}}\right)+c$