Question

Evaluate:
$\int \frac{\sin xdx}{{\sin }^{3}x+{\cos }^{3}x}$

Answer 1

Consider, $I=\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx$

Dividing the numerator and denominator by ${\cos }^{3}x$

$I=\int \frac{\frac{\sin x}{{\cos }^{3}x}}{{\tan }^{3}x+1}dx$

Put $t=\tan x\Rightarrow dt={\sec }^{2}xdx$

$\Rightarrow dx=\frac{dt}{{\sec }^{2}x}$

Substituting the value of $dx$ above we get

$I=\int \frac{t{\sec }^{2}x}{t^3+1}\frac{dt}{{\sec }^{2}x}$

$=\int \frac{t.dt}{t^3+1}$

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$

$=\int \frac{t.dt}{(t+1)(t^2-t+1)}$

Let $\frac{t}{t^2-t+1}=\frac{A}{t+1}+\frac{Bt+C}{t^2-t+1}$

$\Rightarrow \frac{t}{t^2-t+1}=\frac{A(t^2-t+1)+(t+1)(Bt+C)}{(t^2-t+1)(t+1)}$

Simplifying we get :

$t=t^2(A+B)+t(A+B+C)+A+C$

Comparing coefficients we get :

$A+B=0,A+B+C=1,A+C=0$

Solving the three equations simultaneously, we get

$A=\frac{-1}{3},B=C=\frac{1}{3}$

Putting the values of $A,B$ and $C$ our integral becomes,

$\frac{1}{3}\int \frac{t+1}{(t^2-t+1)}dt-\frac{1}{3}\int \frac{dt}{t+1}$

$=\int \frac{1}{6}(\frac{2t-1}{t^2-t+1}+\frac{3}{t^2-t+1})-\frac{\log t+1}{3}$

$=\frac{1}{2}\int \frac{dt}{t^2-t+1}+\frac{\log t^2-t+1}{6}-\frac{\log t+1}{3}$

$=\frac{1}{2}\int \frac{dt}{t^2-2\times t\times \frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1}+\frac{\log t^2-t+1}{6}-\frac{\log t+1}{3}$

$=\frac{1}{2}\int \frac{dt}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+\frac{\log t^2-t+1}{6}-\frac{\log t+1}{3}$

$=\frac{{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\log t^2-t+1}{6}-\frac{\log t+1}{3}+c$

$=\frac{{\tan }^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\log {\tan }^{2}x-\tan x+1}{6}-\frac{\log \tan x+1}{3}+c$ where $t=\tan x$