∫dx/(sin^4x+cos^4x)=

plese solve this.The answer should be in the form of tan

I=∫(1/sin⁴x+cos⁴x)dx

Dividing numerator and denominator by cos⁴x

I=∫(sec⁴x/1+tan⁴x)dx

I=∫[(1+tan²x)sec²x/1+tan⁴x]dx

Now put tanx=t

sec²xdx=dt

I=∫(1+t²/1+t⁴)dt

I=∫[(1+1/t²)/(t²+1/t²)]dt

I=∫[(1+1/t²)/(t-1/t)²+2]dt

Now put (t-1/t)=z

(1+1/t²)dt=dz

I=∫dz/z²+(√2)²=(1/√2)tan-¹z/√2+c

=(1/√2)tan-¹(t-1/t)/√2+c

=(1/√2)tan-¹(tanx-cotx)/√2+c