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What is (xcosx+sinx)dx equal to? (C is an arbitrary constant)
[a] xsinx+C
[b] xcosx+C
[c] -xsinx+C
[d] -xcosx+C

Answer
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Hint: Use the fact that (f(x)+g(x))dx=f(x)dx+g(x)dx and hence prove that the given integral is equal to xcosxdx+sinxdx. In the first integral use integration by parts taking x as first function and cosx as second function as per the ILATE rule. The result of the first function and second function can be related and hence find the value of the given integral.

Complete step-by-step answer:
Let I=(xcosx+sinx)dx
We know that (f(x)+g(x))dx=f(x)dx+g(x)dx(Linearity of integration).
Hence, we have
I=xcosxdx+sinxdx=I1+I2, where I1=xcosxdx and I2=sinxdx
Evaluating I1=xcosxdx
We know that uvdx=uvdx(dudxvdx)dx. This is known as integration by parts rule. Here u is called the first function and v is known as the second function.
While solving questions using integration by parts, it is important to choose the first function and second function in such a way that the integral simplifies. A general order of preference for first function is given by ILATE rule
I = Inverse Trigonometric
L = Logarithmic
A = Algebraic
T = Trigonometric
E = Exponential.
Hence, according to the ILATE rule, we choose u(x) = x and v(x) = cosx.
We have u(x)=cosxdx=sinx and dv(x)dx=ddx(x)=1
Hence, we have
xcosxdx=xsinxsinxdx+C
We have sinxdx=I2
Hence, we have
I1=xsinxI2+C
Adding I2 on both sides, we get
I1+I2=xsinx+C
But we know that I1+I2=I
Hence, we have
I=xsinx+C(xcosx+sinx)dx=xsinx+C

So, the correct answer is “Option A”.

Note: [1] Verification: In case the integrands are simple, we should always verify the correctness of our solution. We can verify the correctness of our solution by checking that the derivative of our answer is equal to the integrand.
We have ddx(xsinx+c)=xddxsinx+sinxddxx+0=xcosx+sinx
Hence, we have
ddx(xsinx+C)=xcosx+sinx
Hence our solution is verified to be correct.