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UPSC NDA II 2015 Mathematics Paper

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Question : 119 of 120

Marks: +1, -0

The value of

\int_{\mathrm{a}}^{\mathrm{b}} \frac{x^{7} + \sin x}{\cos x} \, \mathrm{d}x

where a + b = 0 is

2b – a sin (b - a)
a + 3b cos (b - a)
sin a – (b - a) cos b
0

Solution:

Concept:

Use the property

\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx

to simplify the integral when

a+b=0

.

Explanation:

Let

I = \int_{a}^{b} \frac{x^{7} + \sin x}{\cos x}\,dx

.

Since

a+b=0

, apply the property:

I = \int_{a}^{b} \frac{(a+b-x)^{7} + \sin(a+b-x)}{\cos(a+b-x)}\,dx

.

Because

a+b=0

, we have

a+b-x = -x

.

Thus

I = \int_{a}^{b} \frac{(-x)^{7} + \sin(-x)}{\cos(-x)}\,dx

.

Now

(-x)^{7} = -x^{7}

,

\sin(-x) = -\sin x

, and

\cos(-x) = \cos x

.

So

I = \int_{a}^{b} \frac{-x^{7} - \sin x}{\cos x}\,dx = -\int_{a}^{b} \frac{x^{7} + \sin x}{\cos x}\,dx

.

But the last integral is exactly

I

, so

I = -I

.

Hence

2I = 0

, giving

I = 0

.

Answer:

0

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