What is (e^{ x}+ x) x dx equal to?

Correct Answer is : xsin⁡x+cos⁡x+sin⁡2x2+cx x+ x+{²x}{2}+cxsinx+cosx+2sin2x+c

Correct Answer is : xsin⁡x+cos⁡x+sin⁡2x2+cx x+ x+{²x}{2}+cxsinx+cosx+2sin2x+c

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UPSC NDA I and II 2020 Math Solved Paper

Question : 68 of 120

Marks: +1, -0

What is

\int(e^{\log x}+\sin x)\cos x dx

equal to?

Solution:

Concept:

Use integration by parts and substitution after simplifying

e^{\log x}

x

Explanation:

Let

I = \int (e^{\log x} + \sin x) \cos x \, dx

Since

e^{\log x} = x

, we have

I = \int (x \cos x + \sin x \cos x) \, dx

Separate the integral:

I = \int x \cos x \, dx + \int \sin x \cos x \, dx = I_1 + I_2

For

I_1 = \int x \cos x \, dx

, apply integration by parts with

u = x

dv = \cos x \, dx

Then

du = dx

v = \sin x

I_1 = x \sin x - \int \sin x \, dx = x \sin x + \cos x + c

For

I_2 = \int \sin x \cos x \, dx

, use substitution

t = \sin x

dt = \cos x \, dx

Then

I_2 = \int t \, dt = \frac{t^2}{2} + c = \frac{\sin^2 x}{2} + c

Combine:

I = x \sin x + \cos x + \frac{\sin^2 x}{2} + c

Answer:

Option C:

x \sin x + \cos x + \frac{\sin^{2}x}{2} + c

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