Question

How do I evaluate the integral ∫sin4xcos2xdx?

Answer 1

There's a general method to solve these integrals.

This will work for all formats of integrals having a standard format

1) sin ax cos bx

2) cos ax sin bx

3) cos ax cos bx

4) sin ax sin bx

The method is,

We have

Integration of sin 4x cos 2x

Let me multiply and divide by 2

You'll see why !

1/2 Integration of 2 sin 4x cos 2x

Now, by trig formulas we know something in the form 2 sin ax cos bx can be converted into sums of two sines.

How?

Well,

2 sin ax cos bx = sin ((a+b)x) + sin ((a-b)x)

So,

1/2 Integration of 2 sin 4x cos 2x dx

Becomes,

1/2 Integration of sin 6x + sin 2x

And, we know integration of sin 6x is -1/6 cos 6x and correspondingly, Integration of sin 2x is -1/2 cos 2x

Thus,

Integration of sin 4x cos 2x dx

= -1/12 cos 6x - 1/4 cos 2x + c

Hope this helps you !

Answer 2

Answer:

There's a general method to solve these integrals.

This will work for all formats of integrals having a standard format

1) sin ax cos bx

2) cos ax sin bx

3) cos ax cos bx

4) sin ax sin bx

The method is,

We have

Integration of sin 4x cos 2x

Let me multiply and divide by 2

You'll see why !

1/2 Integration of 2 sin 4x cos 2x

Now, by trig formulas we know something in the form 2 sin ax cos bx can be converted into sums of two sines.

How?

Well,

2 sin ax cos bx = sin ((a+b)x) + sin ((a-b)x)

So,

1/2 Integration of 2 sin 4x cos 2x dx

Becomes,

1/2 Integration of sin 6x + sin 2x

And, we know integration of sin 6x is -1/6 cos 6x and correspondingly, Integration of sin 2x is -1/2 cos 2x

Thus,

Integration of sin 4x cos 2x dx

= -1/12 cos 6x - 1/4 cos 2x + c

Hope this helps you !