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∫dx/(sin^4x+cos^4x)=plese solve this.The answer should be in the form of tan
I=∫(1/sin⁴x+cos⁴x)dxDividing numerator and denominator by cos⁴xI=∫(sec⁴x/1+tan⁴x)dxI=∫[(1+tan²x)sec²x/1+tan⁴x]dxNow put tanx=tsec²xdx=dtI=∫(1+t²/1+t⁴)dtI=∫[(1+1/t²)/(t²+1/t²)]dtI=∫[(1+1/t²)/(t-1/t)²+2]dtNow put (t-1/t)=z(1+1/t²)dt=dzI=∫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
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
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