Question

∫sin2xcos3xdx
details

Answer 1

Answer:

cos5x cosx

- ---------- + ------- + c

10 2

Step-by-step explanation:

To find ---->

------------

∫sin2x cos 3x dx

Solution--->

--------------

I= ∫sin2x cos3x dx

1

= ----- ∫2 cos3x sin2x dx

2

we have a formula from trigonometery as follows

2cosA sinB=sin(A+B)-sin(A-B)

apllying this formula here

1

I =-----∫{sin (3x+2x) - sin(3x-2x)}dx

2

1

=------∫ (sin5x-sinx) dx

2

we have a formula ∫sinx dx=-cosx+c

applying it here

1 -cos5x

=----- { ------------- - (-cosx) } + c

2 5

cos5x cosx

=- ------------- + ---------- + c

10 2

Hope it helps u

Thanks

related other formula--->

-----------------------------------

1:2sinA cosB=sin(A+B) + sin(A-B)

2:2cosA cosB=cos(A+B)+cos(A-B)

3:2sinA sinB =cos((A-B)- cos(A+B)

Answer 2

Answer:

Step-by-step explanation:

∫ sin2x cos3x dx  in this question following formula will be used

2sinA cosB=sin(A+B)+sin(A-B)

ans- 1/2∫ 2sin2x cos3x dx  

     =1/2∫ sin(2x+3x)+sin(2x-3x) dx

     =1/2∫ sin5x+sin(-x) dx        { sin(-θ )= -sinθ }

     =1/2∫ sin5x- sinx dx

     =1/2∫ sin5x dx -1/2 ∫ sinx dx

     =1/10(-cos5x)+1/2cosx (by integration)