Calculate $\int_{0}^{2 π} {cos}^{2} x . d x$ $int_0^(2pi) cos^2x .dx$ ?

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Calculate $\int_{0}^{2 π} {cos}^{2} x . d x$ $int_0^(2pi) cos^2x .dx$ ?

A) Zero
B) 2
C) 1/2
D) $π$ $pi$
The answer is D for your reference

2 Answers

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Answer 1 · 2017-09-05T16:45:06

Yes, the correct answer is $D$ $D$ , or $π$ $pi$ .

Explanation:

Use the power reduction formula

${cos}^{2} x = \frac{1 + cos 2 x}{2}$ $cos^2x= (1 + cos2x)/2$ .

So we have:

$I = \int_{0}^{2 π} \frac{1 + cos 2 x}{2}$ $I = int_0^(2pi) (1 + cos2x)/2$

$I = \frac{1}{2} \int_{0}^{2 π} 1 + cos (2 x) d x$ $I = 1/2int_0^(2pi) 1 + cos(2x)dx$

$I = \frac{1}{2} \int_{0}^{2 π} 1 d x + \frac{1}{2} \int_{0}^{2 π} cos (2 x) d x$ $I = 1/2int_0^(2pi) 1 dx + 1/2int_0^(2pi) cos(2x)dx$

If we let $u = 2 x$ $u = 2x$ in the second integral, then $\frac{1}{2} d u = d x$ $1/2du = dx$ . The integral of $cos u$ $cosu$ is $sin u$ $sinu$ and reverse your substitution to get $sin (2 x)$ $sin(2x)$ .

$I = \frac{1}{2} {[x]}_{0}^{2 π} + \frac{1}{2} (\frac{1}{2}) {[sin (2 x)]}_{0}^{2 π}$ $I = 1/2[x]_0^(2pi) + 1/2(1/2)[sin(2x)]_0^(2pi)$

The integral equals

$I = \frac{1}{2} (2 π) + \frac{1}{4} (sin (4 π) - sin (0))$ $I = 1/2(2pi) + 1/4(sin(4pi) - sin(0))$

$I = π - \frac{1}{4} (0 - 0)$ $I = pi - 1/4(0 - 0)$

$I = π$ $I = pi$

Answer $D$ $D$ , as required.

Hopefully this helps!

Answer 2 · 2017-09-06T11:11:15

$\int_{0}^{2 π} {cos}^{2} (x) d x$ $int_0^(2pi)cos^2(x)dx$

We can rewrite this:

$\int_{0}^{2 π} {cos}^{2} (x) d x = \int_{0}^{2 π} ({cos}^{2} (x) - {sin}^{2} (x) + {sin}^{2} (x)) d x$ $int_0^(2pi)cos^2(x)dx=int_0^(2pi)(cos^2(x)-sin^2(x)+sin^2(x))dx$

Note that $cos (2 x) = {cos}^{2} (x) - {sin}^{2} (x)$ $cos(2x)=cos^2(x)-sin^2(x)$ :

$\int_{0}^{2 π} {cos}^{2} (x) d x = \int_{0}^{2 π} cos (2 x) d x + \int_{0}^{2 π} {sin}^{2} (x) d x$ $int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)sin^2(x)dx$

Rewrite ${sin}^{2} (x)$ $sin^2(x)$ using ${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x)=1$ :

$\int_{0}^{2 π} {cos}^{2} (x) d x = \int_{0}^{2 π} cos (2 x) d x + \int_{0}^{2 π} (1 - {cos}^{2} (x)) d x$ $int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)(1-cos^2(x))dx$

$\int_{0}^{2 π} {cos}^{2} (x) d x = \int_{0}^{2 π} cos (2 x) d x + \int_{0}^{2 π} d x - \int_{0}^{2 π} {cos}^{2} (x) d x$ $int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)dx-int_0^(2pi)cos^2(x)dx$

Add $\int_{0}^{2 π} {cos}^{2} (x) d x$ $int_0^(2pi)cos^2(x)dx$ to both sides of the equation:

$2 \int_{0}^{2 π} {cos}^{2} (x) d x = \int_{0}^{2 π} cos (2 x) d x + \int_{0}^{2 π} d x$ $2int_0^(2pi)cos^2(x)dx=int_0^(2pi)cos(2x)dx+int_0^(2pi)dx$

Evaluate these integrals:

$2 \int_{0}^{2 π} {cos}^{2} (x) d x = {∣ ∣ ∣ \frac{1}{2} sin (2 x) ∣ ∣ ∣}_{0}^{2 π} + x {∣ ∣ ∣}_{0}^{2 π}$ $2int_0^(2pi)cos^2(x)dx=|1/2sin(2x)|_0^(2pi) +x|_0^(2pi)$

$2 \int_{0}^{2 π} {cos}^{2} (x) d x = \frac{1}{2} (sin (4 π) - sin (0)) + (2 π - 0)$ $2int_0^(2pi)cos^2(x)dx=1/2(sin(4pi)-sin(0))+(2pi-0)$

$2 \int_{0}^{2 π} {cos}^{2} (x) d x = 2 π$ $2int_0^(2pi)cos^2(x)dx=2pi$

Dividing by $2$ $2$ :

$\int_{0}^{2 π} {cos}^{2} (x) d x = π$ $int_0^(2pi)cos^2(x)dx=pi$

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Calculate $\int_{0}^{2 π} {cos}^{2} x . d x$ $int_0^(2pi) cos^2x .dx$ ?

A) Zero
B) 2
C) 1/2
D) $π$ $pi$
The answer is D for your reference

2 Answers

Explanation:

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Explanation:

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Impact of this question

Calculate $\int_{0}^{2 π} {cos}^{2} x . d x$ $int_0^(2pi) cos^2x .dx$ ?

A) Zero
B) 2
C) 1/2
D) $π$ $pi$
The answer is D for your reference