What is e^{ x} \; x³ x - x/² x dx [NDA II 2016]

Correct Answer is : (x−sec⁡x)esin⁡x+c(x - x) e^{ x} + c(x−secx)esinx+c

Correct Answer is : (x−sec⁡x)esin⁡x+c(x - x) e^{ x} + c(x−secx)esinx+c

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UPSC NDA 2 2016 Mathematics Solved Paper

Question : 98 of 120

Marks: +1, -0

What is

\int e^{\sin x} \; \frac{x\cos^3 x - \sin x}{\cos^2 x} dx

Solution:

Let us differentiate all the options one by one to get the expression in the question whose integral is to be found.

Here

x e^{\sin x}

is the common term in all the options. So,let us differentiate it first.

Let

l = x e^{\sin x}

\Rightarrow \; \frac{dl}{dx} = e^{\sin x} [x \cos x + 1]

\Rightarrow \; \frac{dl}{dx} = \; \frac{e^{\sin x}}{\cos^2 x} [x \cos^3 x + \cos^2 x]

Let

m = \sec x e^{\sin x}

\Rightarrow \; \frac{dm}{dx} = \sec x e^{\sin x} \cdot \cos x + e^{\sin x} \sec x \tan x

\Rightarrow \; \frac{dm}{dx} = e^{\sin x} \left[1 + \; \frac{\sin x}{\cos^2 x}\right]

\Rightarrow \; \frac{dm}{dx} = \; \frac{e^{\sin x}}{\cos^2 x} [\cos^2 x + \sin x]

Differentiation of option (a) is

= \; \frac{e^{\sin x}}{\cos^2 x} [x \cos^3 x + \cos^2 x + \cos^2 x + \sin x]

= \; \frac{e^{\sin x}}{\cos^2 x} [x \cos^3 x + 2 \cos^2 x + \sin x]

Differentiation of option (b) is

= \; \frac{e^{\sin x}}{\cos^2 x} [x \cos^3 x + \cos^2 x - \cos^2 x - \sin x]

= \; \frac{e^{\sin x}}{\cos^2 x} [x \cos^3 x - \sin x]

∴ Option (b) is correct.

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