Question

integrate the following:
$\sin ^{2} (2x + 5)$
​

Answer 1

Answer:

(1/2) x - (1/8) Sin (4x + 10 ) + c

Step-by-step explanation:

To find ---> ∫ Sin² (2x + 5 ) dx

-----------

Solution--->

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

We have a formula

Cos2θ = 1 - 2 Sin²θ

=> 2 Sin²θ = 1 - Cos 2θ

Putting θ = 2x + 5

=> 2 Sin²(2x + 5) = 1 - Cos 2(2x+5)

=> 2 Sin² (2x + 5)= 1 - Cos (4x + 10)

Now,

I = ∫ Sin² (2x + 5) dx

= 1/2 ∫ 2 Sin² (2x + 5) dx

= 1/2 ∫ {1 - Cos( 4x + 10 ) } dx

= 1/2 ∫ 1 dx - 1/2 ∫ Cos (4x + 10) dx

We know that

∫ 1 dx = x + c

∫ Cosx dx = sinx + c

Applying these formulee

I= ( 1/2 )x - 1/2 Sin (4x +10) / 4 +c

= (1/2) x - (1/8 ) Sin (4x + 10 ) + c