Question

Integrate cos^2(2x+1) with full steps plzzzzzz......Plzz help......answer only if u know plzz​

Answer 1

Answer:

$\large\bold\red{\frac{ \sin2(2x + 1) }{8} + \frac{2x + 1}{4} + c}$

Step-by-step explanation:

Given,

$\int { \cos }^{2} (2x + 1)dx$

Let,

$2x + 1 = y \\ \\ = > 2dx = dy \\ \\ = > dx = \frac{dy}{2}$

Therefore,

putting the value,

we get,

$= \frac{1}{2} \int { \cos }^{2} y \: dy$

Now,

we know that,

$\cos(2 \alpha ) = 2 { \cos}^{2} \alpha - 1 \\ \\ = > 2 { \cos }^{2} \alpha = 1 + \cos(2 \alpha ) \\ \\ = > { \cos }^{2} \alpha = \frac{1 + \cos(2 \alpha ) }{2}$

Therefore,

from this formula,

we get,

$= \frac{1}{2} \times \frac{1}{2} \int( \cos2y + 1)dy \\ \\ = \frac{1}{4} \int( \cos2y + 1)dy \\ \\ = \frac{1}{4} \times ( \frac{ \sin2y }{2} + y) + c \\ \\ = \frac{ \sin2y }{8} + \frac{y}{4} + c$

Putting the value of y =( 2x +1) ,

we get,

$=\large \bold{\frac{ \sin2(2x + 1) }{8} + \frac{2x + 1}{4} + c}$

Where,

C is integral constant

Answer 2

Answer:

deleted again.....who is this.......