Question

Solve:$\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx$

Answer 1

$I=\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx=\int \frac{\sin x}{(\sin x+\cos x)({\sin }^{2}x-\sin x\cos x+{\cos }^{2}x)}dx[\because a^3+b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]=\int \frac{\sin x}{(\sin x+\cos x)(1-\sin x\cos x)}dx=\int \frac{2\sin x}{(\sin x+\cos x)(2-2\sin x\cos x)}dx=\int \frac{2\sin x+\cos x-\cos x}{(\sin x+\cos x)(2-2\sin x\cos x)}dx=\int \frac{\sin x+\cos x}{(\sin x+\cos x)(2-2\sin x\cos x)}dx+\int \frac{\sin x-\cos x}{(\sin x+\cos x)(2-2\sin x\cos x)}dx=\int \frac{1}{(2-2\sin x\cos x)}dx+\int \frac{\sin x-\cos x}{(\sin x+\cos x)(2-2\sin x\cos x)}dx$

Let,

$I_1=\int \frac{1}{(2-2\sin x\cos x)}dx,I_2=\int \frac{\sin x-\cos x}{(\sin x+\cos x)(2-2\sin x\cos x)}dx$

First solving $I_1$ we get,

$I_1=\int \frac{1}{(2-2\sin x\cos x)}dx=\int \frac{1}{2-2\tan x{\cos }^{2}x}dx=\int \frac{1}{2-2\frac{\tan x}{{\sec }^{2}x}}dx=\int \frac{{\sec }^{2}x}{2{\sec }^{2}x-2\tan x}dx=\int \frac{{\sec }^{2}x}{2{\tan }^{2}x+2-2\tan x}dx$

Let,

$y=\tan x\Rightarrow \frac{dy}{dx}={\sec }^{2}x\Rightarrow dy={\sec }^{2}xdx$

$\therefore I_1=\int \frac{1}{2y^2+2-2y}dy=\int \frac{1}{2y^2-2y+\frac{1}{2}-\frac{1}{2}+2}dy=\int \frac{1}{(\sqrt{2}y-\frac{1}{\sqrt{2}}{)}^2-\frac{3}{2}}dy$

Let us assume,

$\sqrt{2}y-\frac{1}{\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\tan z\Rightarrow \sqrt{2}dy=\frac{\sqrt{3}}{\sqrt{2}}{\sec }^{2}zdz$

Substituting these values in $I_1$,

$I_1=\int \frac{\sqrt{3}{\sec }^{2}z}{2(\frac{3}{2}{\tan }^{2}z-\frac{3}{2})}dz=\int \frac{\sqrt{3}{\sec }^{2}z}{3{\sec }^{2}z}dz=\int \frac{1}{\sqrt{3}}dz=\frac{z}{\sqrt{3}}$

Substituting $z$ in terms of $y$ we get,

$I_1=\frac{\arctan (\frac{2y-1}{\sqrt{3}})}{\sqrt{3}}$

Substituting $y$ in terms of $x$ we get,

$I_1=\frac{\arctan \left(\frac{2\tan x-1}{\sqrt{3}}\right)}{\sqrt{3}}$

Again, solving $I_2$ we get,

$I_2=\int \frac{\sin x-\cos x}{(\sin x+\cos x)(2-2\sin x\cos x)}dx=\int \frac{(\sin x-\cos x)(\sin x+\cos x)}{{(\sin x+\cos x)}^2(2-2\sin x\cos x)}dx=\int \frac{{\sin }^{2}x-{\cos }^{2}x}{({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x)(2-2\sin x\cos x)}dx=\int \frac{-\cos 2x}{(1+\sin 2x)(2-\sin 2x)}dx=\int \frac{-(1+2)\cos 2x}{3(1+\sin 2x)(2-\sin 2x)}dx=\int \frac{-(1+\sin 2x+2-\sin 2x)\cos 2x}{3(1+\sin 2x)(2-\sin 2x)}dx=-\int \frac{(1+\sin 2x)\cos 2x}{3(1+\sin 2x)(2-\sin 2x)}dx-\int \frac{(2-\sin 2x)\cos 2x}{3(1+\sin 2x)(2-\sin 2x)}dx=-\int \frac{\cos 2x}{3(2-\sin 2x)}dx-\int \frac{\cos 2x}{3(1+\sin 2x)}dx=\int \frac{1}{3(2-\sin 2x)}d(2-\sin 2x)-\int \frac{1}{3(1+\sin 2x)}d(1+\sin 2x)=\frac{1}{3}\ln (2-\sin 2x)-\frac{1}{3}\ln (1+\sin 2x)\therefore I=\frac{\arctan \left(\frac{2\tan x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{1}{3}\ln (2-\sin 2x)-\frac{1}{3}\ln (1+\sin 2x)\therefore \int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx=\frac{\arctan \left(\frac{2\tan x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{1}{3}\ln (2-\sin 2x)-\frac{1}{3}\ln (1+\sin 2x).$