Math, asked by mvgreddykdp, 10 months ago

sin^ 4 A -cos^4 A +1​

Answers

Answered by shardadevi99554
0

Step-by-step explanation:

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. { { {3 {5 {2}^{?} }^{?} }^{2} }^{2} }^{2}  {?}^{?}

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Answered by vickey90
0

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How to integrate dx/(sin(x)+sin(2x)) ?

Calculus

1 Answer

Noah G · maganbhai P.

Feb 23, 2018

I

=

1

2

ln

|

cos

x

+

1

|

+

1

6

ln

|

cos

x

1

|

2

3

ln

|

2

cos

x

+

1

|

+

C

Explanation:

Using some trig:

I

=

1

sin

x

+

2

sin

x

cos

x

d

x

I

=

1

sin

x

(

1

+

2

cos

x

)

d

x

I

=

sin

x

sin

2

x

(

1

+

2

cos

x

)

d

x

I

=

sin

x

(

cos

2

x

1

)

(

1

+

2

cos

x

)

d

x

I

=

sin

x

(

cos

x

+

1

)

(

cos

x

1

)

(

2

cos

x

+

1

)

d

x

Now we let

u

=

cos

x

. Then

d

u

=

sin

x

d

x

and

d

x

=

d

u

sin

x

.

I

=

sin

x

(

cos

x

+

1

)

(

cos

x

1

)

(

2

cos

x

+

1

)

d

u

sin

x

I

=

1

(

u

+

1

)

(

u

1

)

(

2

u

+

1

)

d

u

This becomes a partial fraction problem.

A

u

+

1

+

B

u

1

+

C

2

u

+

1

=

1

(

u

+

1

)

(

u

1

)

(

2

u

+

1

)

A

(

u

1

)

(

2

u

+

1

)

+

B

(

u

+

1

)

(

2

u

+

1

)

+

C

(

u

+

1

)

(

u

1

)

=

1

A

(

2

u

2

2

u

+

u

1

)

+

B

(

2

u

2

+

2

u

+

u

+

1

)

+

C

(

u

2

1

)

=

1

A

(

2

u

2

u

1

)

+

B

(

2

u

2

+

3

u

+

1

)

+

C

(

u

2

1

)

=

1

2

A

u

2

A

u

A

+

2

B

u

2

+

3

B

u

+

B

+

C

u

2

C

=

1

Thus

2

A

+

2

B

+

C

=

0

3

B

A

=

0

B

A

C

=

1

If we substitute the second equation

A

=

3

B

into the first and third we get.

3...

B

(

3

B

)

C

=

1

2

B

C

=

1

1...

2

(

3

B

)

+

2

B

+

C

=

0

6

B

+

2

B

+

C

=

0

8

B

+

C

=

0

We can easily solve this system by elimination .

6

B

=

1

B

=

1

6

Therefore

8

(

1

6

)

+

C

=

0

C

=

4

3

And

A

=

3

B

=

3

(

1

6

)

=

1

2

Thus the partial fraction decomposition is as follows:

1

2

(

u

+

1

)

+

1

6

(

u

1

)

4

3

(

2

u

+

1

)

I

=

1

2

1

u

+

1

d

u

+

1

6

1

u

1

d

u

2

3

2

(

2

u

)

+

1

d

u

This can be easily integrated.

I

=

1

2

ln

|

u

+

1

|

+

1

6

ln

|

u

1

|

2

3

ln

|

2

u

+

1

|

+

C

subst. back ,

u

=

cos

x

I

=

1

2

ln

|

cos

x

+

1

|

+

1

6

ln

|

cos

x

1

|

2

3

ln

|

2

cos

x

+

1

|

+

C

Hopefully this helps!

Note:

A

,

B

,

C

can obtain by putting

u

=

1

,

u

=

1

and

u

=

1

2

into

BLUE EQUATION :

A

(

u

1

)

(

2

u

+

1

)

+

B

(

u

+

1

)

(

2

u

+

1

)

+

C

(

u

+

1

)

(

u

1

)

=

1

u

=

1

A

(

1

1

)

(

2

+

1

)

+

B

(

0

)

+

C

(

0

)

=

1

A

(

2

)

(

1

)

=

1

A

=

1

2

u

=

1

A

(

0

)

+

B

(

1

+

1

)

(

2

+

1

)

+

C

(

0

)

=

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