Integeration help..... substituiton or triginometric identites method

1) $sin4x/cos2x
2) $xsquare + 1/x to the power 4 + 1dx
2$dx/2sinx+cosx+3



by substititon and trigonometric method only!! please ans fast!!!

If your second question is ( x² + 1 ) / ( x⁴ + 1 ), then,

I am afraid, the above answer is not correct.

In that case, proceed as follows :

... ∫ [ ( x² + 1 ) / ( x⁴ + 1 ) ] dx ............ Divide Num & Den by x²

= ∫ { [ 1 + ( 1/x² ) ] / [ x² + ( 1/x² ) ] } dx

= ∫ { 1 / [ ( x² + (1/x²) - 2 ) + 2 ] } • [ 1 + ( 1/x² ) ] dx

= ∫ { 1 / [ (x - (1/x))² + 2 ] } • [ 1 - ( -1/x² ) ] dx

= ∫ { 1 / [ u² + (√2)² ] } du, ...... u = x - (1/x), du = [ 1 - (-1/x²) ] dx

= (1/√2). tanֿ¹ (u/√2) + C

= (1/√2). tanֿ¹ [ (1/√2)( x - (1/x)) ] + C

= (1/√2). tanֿ¹ [ ( x² - 1 ) / (√2.x) ] + C ............... Ans.
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You are welcome, joe !

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.Q.Intergrate the following :
a) ∫ (sin(4x)/cos(2x) ) dx
b) ∫ {x^2+(1/x)^4+1} dx
c) ∫{1 / [2sin(x)+cos(x)+3] } dx

Ans : a) ∫ (sin(4x)/cos(2x) ) dx
Assume : I = ∫ (sin(4x)/cos(2x) ) dx

Apply : sin(4x) = 2*sin(2x)*cos(2x)
Therefore : I = ∫ ([2*sin(2x)*cos(2x)]/cos(2x) ) dx
I =2* ∫sin(2x) dx

Apply : ∫sin(nx) dx = -cos(nx) / n
Therefore : I =2* {-cos(2x) / 2} +C
=> I = -cos(2x) +C

b) ∫ {x^2+(1/x)^4+1} dx
Assume : I = ∫ {x^2+(1/x)^4+1} dx
=> I = ∫ (x^2) dx+∫ (1/x)^4 dx+∫ 1 dx
=> I = ∫ {x^2} dx+∫ x^(-4) dx+∫ (1) dx

Apply : ∫ x^n dx =[ x^(n+1) ] / (n+1)
=> I ={ [x^3] / 3 }+ { [x^(-3)] /-3}+ {[x]} +C
=> I ={ [x^3] / 3 }- { [x^(-3)] /3}+ x +C
=> I ={ [x^3] / 3 }- {1 / [3*x^(3)] }+ x +C

c) ∫{1 / [2sin(x)+cos(x)+3] } dx
Assume : I = ∫{1 / [2sin(x)+cos(x)+3] } dx ---------(1)