Question

integral(sin^2x +cos^4x dx)​

Answer 1

Answer:

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Answer 2

∫

s

i

n

2

x

c

o

s

4

x

d

x

∫sin2xcos4xdx

∫

(

1

−

c

o

s

2

x

)

(

c

o

s

4

x

)

∫(1−cos2x)(cos4x)

∫

(

c

o

s

4

x

−

c

o

s

6

x

)

d

x

∫(cos4x−cos6x)dx

cos

2

x

=

1

+

cos

2

x

2

cos2⁡x=1+cos⁡2x2

1

2

∫

c

o

s

4

x

d

x

−

1

2

∫

c

o

s

6

x

d

x

12∫cos4xdx−12∫cos6xdx

1

8

∫

(

1

+

c

o

s

2

x

)

2

−

1

16

∫

(

1

+

c

o

s

2

x

)

3

18∫(1+cos2x)2−116∫(1+cos2x)3