The integral of xcosx is equal to xsinx + cosx + C, where C is the constant of integration. We can evaluate the integration of xcosx using the integration by parts method of integration. The integral of a function is nothing but its antiderivative as integration is the reverse process of differentiation. Hence, the integral of xcosx is equal to the antiderivative of xcosx which is written mathematically as, ∫xcosx dx = xsinx + cosx + C.
In this article, let us calculate the integral of xcosx and derive its formula using integration by parts method. We will also evaluate the definite integration of xcosx with limits ranging from 0 to pi/2 and solve some examples using the integral of xcosx for a better understanding of the concept.
| 1. | What is the Integral of xcosx? |
| 2. | Integral of xcosx Formula |
| 3. | Integral of xcosx Proof |
| 4. | Definite Integration of xcosx |
| 5. | FAQs on Integral of xcosx |
The integral of xcosx gives the area under the curve of the function f(x) = xcosx and gives different equivalent answers when evaluated using different methods of integration. The integration of xcosx is given by, ∫xcosx dx = xsinx + cosx + C, where C is the integration constant, ∫ is the symbol of integration and dx shows the integration of xcosx is with respect to x. In the next section, let us go through the formula for the integral of xcosx.
The formula for the integral of xcosx is given by, ∫xcosx dx = xsinx + cosx + C, where C is the constant of integration. We can determine this integral of xcosx using the integration by parts method of integration. The formula for the integral of xcosx is shown in the image given below:
Now that we know that the intergal of xcosx is equal to xsinx + cosx + C, we will prove it using the integration by parts method of interaction. The formula for integration by parts is given by, ∫udv = uv - ∫vdu. For the intergal of xcosx, that is, ∫xcosx dx, assume u = x and dv = cosx dx, then v = sinx and du = dx because the derivative of sinx is equal to cosx. Therefore, using the given formula, we have
= xsinx - (-cosx) + C ---- [Because the integral of sinx is equal to -cosx + C]
= xsinx + cosx + C, where C is the integration constant.
Hence, we have proved that the integral of xcosx is equal to xsinx + cosx + C.
In this section, we will use the formula for the integration of xcosx which is given by ∫xcosx dx = xsinx + cosx + C to determine the definite integral of xcosx with limits ranging from 0 to π/2. Using the formula, we have
Hence, the value of the definite integral of xcosx with limits from 0 to pi/2 is equal to π/2 - 1.
Important Notes on Integral of xcosx
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Example 1: What is the integral of xcosx with limits from -π/2 to π/2? Solution: To find the integral of xcosx from -π/2 to π/2, we will use its formula given by, ∫xcosx dx = xsinx + cosx + C. \(\begin\int_<-\frac<\pi>>^<\frac<\pi>>x\cos xdx&=\left [ x \sin x+\cos x+ C \right ]_<-\frac<\pi>>^<\frac<\pi>>\\&=(\frac<\pi>\sin (\frac<\pi>)+\cos \frac<\pi>+C)-((-\frac<\pi>)\sin (-\frac<\pi>)+\cos (-\frac<\pi>)+C)\\&=\frac<\pi>\sin (\frac<\pi>)+\cos \frac<\pi>+C-\frac<\pi>\sin (\frac<\pi>)-\cos (-\frac<\pi>)-C\\&=\frac<\pi>+0-\frac<\pi>-0\\&=0\end\) Answer: Integral of xcosx from -π/2 to π/2 is equal to 0.
Example 2: Calculate the integral of xsinx. Solution: We will evaluate the integral of xsinx in the same way as we determined the integral of xcosx. We will use the method of integration by parts. The formula for integration by parts is given by, ∫udv = uv - ∫vdu. For the intergal of xsinx, that is, ∫xsinx dx, assume u = x and dv = sinx dx, then v = -cosx and du = dx because the derivative of cosx is equal to -sinx. ∫xsinx dx = ∫udv = uv - ∫v du = -xcosx + ∫cosx dx = -xcosx + sinx + C ---- [Because the integral of cosx is equal to sinx + C] = -xcosx + sinx + C, where C is the integration constant. Answer: ∫xsinx dx = -xcosx + sinx + C
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