Question

`int 1/(1-sin^4(x)) dx=`?

Answer 1

`1/2tanx-1/(2sqrt2)*arc tan(1/sqrt2*cotx)+C`.

Explanation:

Let, `I=int1/(1-sin^4x)dx`.

`:. I=1/2int2/{(1-sin^2x)(1+sin^2x)}dx`,

`=1/2int{(1-sin^2x)+(1+sin^2x)}/{(1-sin^2x)(1+sin^2x)}dx`,

`=1/2int{(1-sin^2x)/((1-sin^2x)(1+sin^2x))+(1+sin^2x)/((1-sin^2x)(1+sin^2x))}dx`,

`=1/2int{1/(1+sin^2x)+1/(1-sin^2x)}dx`,

`=1/2I_1+1/2int1/cos^2xdx`,

`=1/2I_1+1/2intsec^2xdx`,

`=1/2I_1+1/2tanx," where, "`

`I_1=int1/(1+sin^2x)dx`,

`=int1/{sin^2x(csc^2x+1)}dx`,

`=intcsc^2x/{(cot^2x+1)+1}dx`,

`=intcsc^2x/(cot^2x+2)dx`,

`=-int(-csc^2x)/(cot^2x+2)dx`,

`=-intdy/{y^2+(sqrt2)^2},...[because, cotx=y rArr -csc^2xdx=dy]`,

`=-1/sqrt2*arc tan(y/sqrt2)`,

`=-1/sqrt2*arc tan(1/sqrt2*cotx)`.

Altogether, we have,

`I=1/2tanx-1/(2sqrt2)*arc tan(1/sqrt2*cotx)+C`.

Feel the Joy of Maths.!

Answer 2

The answer is `=1/2tanx+1/(2sqrt2)arctan(sqrt2tanx)+C`

Explanation:

Perform the substitution

`sinx=tanx/secx`

`sec^2x=1+tan^2x`

Therefore,

`1/(1-sin^4x)=1/(1-(tan^4x/sec^4x))=(sec^4x)/(sec^4x-tan^4x)`

`=(sec^2x(1+tan^2x))/((1+tan^2x)^2-tan^4x)`

`=(sec^2x(1+tan^2x))/(1+2tan^2x)`

Then,

Let `u=tanx`, `=>`, `du=sec^2xdx`

Therefore the integral is

`I=int(dx)/(1-sin^4x)=int(sec^2x(1+tan^2x)dx)/(1+2tan^2x)`

`=int((1+u^2)du)/(1+2u^2)`

`=int((1/2(1+2u^2)+1/2)du)/(1+2u^2)`

`=1/2intdu+1/2int(1du)/(1+2u^2)`

`=1/2u+1/2int(1du)/(1+2u^2)`

Let `v=sqrt2u`, `=>`, `dv=sqrt2du`

`1/2int(1du)/(1+2u^2)=1/(2sqrt2)int(dv)/(1+v^2)`

`=1/(2sqrt2)arctanv`

`=1/(2sqrt2)arctan(sqrt2u)`

`=1/(2sqrt2)arctan(sqrt2tanx)`

Putting it all together

`I=1/2tanx+1/(2sqrt2)arctan(sqrt2tanx)+C`