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Math  /  Calculus

QuestionUse integration tables to evaluate the integral ∫0π227xcos⁡xdx\int_{0}^{\frac{\pi}{2}} 27 x \cos x d x.

Studdy Solution

STEP 1

1. The integral to be evaluated is ∫0π227xcos⁡x dx\int_{0}^{\frac{\pi}{2}} 27 x \cos x \, dx.
2. We can use integration tables to find the antiderivative of the integrand.
3. We will apply the limits of integration to the antiderivative to find the definite integral.

STEP 2

1. Identify and use the appropriate integration formula from integration tables.
2. Find the antiderivative of 27xcos⁡x27 x \cos x.
3. Evaluate the antiderivative at the upper limit π2\frac{\pi}{2}.
4. Evaluate the antiderivative at the lower limit 00.
5. Subtract the value of the antiderivative at the lower limit from the value at the upper limit.

STEP 3

Identify the appropriate integration formula from the integration tables. The integral ∫xcos⁡x dx\int x \cos x \, dx can be solved using integration by parts. The formula for integration by parts is: ∫udv=uv−∫vdu \int u dv = uv - \int v du For u=xu = x and dv=cos⁡x dxdv = \cos x \, dx, we have du=dxdu = dx and v=sin⁡xv = \sin x. Therefore, ∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx \int x \cos x \, dx = x \sin x - \int \sin x \, dx

STEP 4

Continue with the integration by parts to find the antiderivative: ∫xcos⁡x dx=xsin⁡x−(−cos⁡x) \int x \cos x \, dx = x \sin x - (-\cos x) ∫xcos⁡x dx=xsin⁡x+cos⁡x \int x \cos x \, dx = x \sin x + \cos x Multiply by the constant 2727: ∫27xcos⁡x dx=27(xsin⁡x+cos⁡x) \int 27 x \cos x \, dx = 27 \left( x \sin x + \cos x \right)

STEP 5

Evaluate the antiderivative at the upper limit π2\frac{\pi}{2}: 27(xsin⁡x+cos⁡x)∣0π2=27((π2sin⁡π2+cos⁡π2)−(0⋅sin⁡0+cos⁡0)) 27 \left( \left. x \sin x + \cos x \right) \right|_{0}^{\frac{\pi}{2}} = 27 \left( \left( \frac{\pi}{2} \sin \frac{\pi}{2} + \cos \frac{\pi}{2} \right) - \left( 0 \cdot \sin 0 + \cos 0 \right) \right)

STEP 6

Simplify the expression at the upper limit π2\frac{\pi}{2}: π2sin⁡π2+cos⁡π2=π2⋅1+0=π2 \frac{\pi}{2} \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = \frac{\pi}{2} \cdot 1 + 0 = \frac{\pi}{2}

STEP 7

Simplify the expression at the lower limit 00: 0⋅sin⁡0+cos⁡0=0+1=1 0 \cdot \sin 0 + \cos 0 = 0 + 1 = 1

STEP 8

Subtract the value at the lower limit from the value at the upper limit: 27(π2−1) 27 \left( \frac{\pi}{2} - 1 \right) =27⋅π2−27⋅1 = 27 \cdot \frac{\pi}{2} - 27 \cdot 1 =27π2−27 = \frac{27\pi}{2} - 27
Solution: The evaluated integral is: ∫0π227xcos⁡x dx=27π2−27 \int_{0}^{\frac{\pi}{2}} 27 x \cos x \, dx = \frac{27\pi}{2} - 27

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