Evaluate the following : $\begin{align*} \int \frac{\sin^6x+\cos^6x}{\sin^2x \cos^2x}dx \end{align*}$

Answer 1

Toolbox:

• $\sin ^2 x+ \cos ^2 x=1$
• $\int \sec^2x dx=\tan x+c$
• $\int cosec^2 xdx=-\cot x+c$

Let $\large\int\frac{\sin^6x+\cos^6x}{sin^2x\cos^2x}dx$

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

Similarly $(\sin ^2 x+\cos ^2 x)^3-3 \sin ^2 x\cos ^2 x(\sin ^2 x+\cos ^2 x)$

$=\sin ^6x+\cos ^6x$

Therefore $I=\large\int \frac {(\sin ^2 x+\cos ^2 x)^3-3 \sin ^2 x\cos ^2 x(\sin ^2 x+\cos ^2 x)}{\sin ^2x \cos ^2 x}dx$

But $\sin ^2 x+\cos ^2 x=1$

Therefore $I=\large\int \frac{1-3 \sin ^2 x\cos ^2 x}{\sin ^2x \cos ^2 x}dx$

On spliting the terms we get,

$I=\int \large \frac {1}{\sin ^2x \cos ^2 x}dx-\int \frac{3 \sin ^2 x\cos ^2 x}{\sin ^2 x\cos ^2 x}dx$

$=\large\int \frac{\sin ^2 x+\cos ^2x}{\sin ^2x\cos ^2x}dx-\int 3 dx$

$=\large\int \frac{\sin ^2 x}{\sin ^2 x \cos ^2 x}dx+\int \frac{\cos ^2 x}{\sin ^2 x\cos^2 x}dx-\int 3 dx$

$=\int (\sec^2 x dx+\int cosec ^2 x dx-\int 3 dx)$

On integrating we get

$\tan x -\cot x -3x+c$