Question

Evaluate ∫cos2 xdx_________

• A. -x/2+sin2x/4+c
• B. x/2+sin2x/2+c
• C. x/2+sin2x/4+c
• D. -x/2+cos4x/4+c

Answer 1

The Correct Option is A

To evaluate the integral ∫cos²xdx, we can use the trigonometric identity for cos²x, which is:

cos²x = (1 + cos2x) / 2

Substituting this identity into the integral, we have:

∫cos²xdx = ∫(1 + cos2x)/2 dx

We can split the integral into two separate parts:

= (1/2)∫dx + (1/2)∫cos2xdx

The first integral is straightforward:

(1/2)∫dx = x/2

The second integral requires substituting 2x with a new variable, u. Let:

u = 2x, then du = 2dx, or dx = du/2

Now substitute into the integral:

(1/2)∫cos2xdx = (1/2)∫cosudu/2 = (1/4)∫cosudu

The integral of cosu is sinu, so we have:

(1/4)∫cosudu = (1/4)sinu + C

Re-substitute back for u = 2x:

(1/4)sin2x + C

Combine the results from both integrals:

∫cos²xdx = x/2 + sin2x/4 + C

Therefore, the evaluated integral is:

-x/2 + sin2x/4 + c