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AP EAMCET Engineering 17 Sep 2020 Shift 2 Paper

Section: Mathematics

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Question : 11 of 160

Marks: +1, -0

If

\int \frac{\cos(13x)-\cos(14x)}{1+2\cos(9x)} \, dx = \frac{\sin(4x)}{a} - \frac{\sin(5x)}{b} + c,

a^{b}=

$4^{5}$
$5^{4}$
$4^{4}$
$5^{5}$

Solution:

Given,

I=\int\; \frac{\cos 13x - \cos 14x}{1+2\cos 9x} dx

Let,

\;\; f(x)=\; \frac{\cos 13x - \cos 14x}{1+2\cos 9x}

=\; \frac{\sin 9x (\cos 13x - \cos 14x)}{\sin 9x + \sin 18x}

=\; \frac{\sin 9x - 2 \sin\frac{27}{2}x \cdot \sin\left(-\frac{x}{2}\right)}{2 \sin\frac{27x}{2} \cdot \cos\left(-\frac{9x}{2}\right)}

=\; \frac{\sin 9x \cdot \sin\frac{x}{2}}{\cos\left(\frac{9x}{2}\right)} = 2 \sin\frac{9x}{2} \cdot \sin\frac{x}{2}

= \cos\left(\frac{9x}{2} - \frac{x}{2}\right) - \cos\left(\frac{9x}{2} + \frac{x}{2}\right)

= \cos 4x - \cos 5x

So,

\;\; I=\int (\cos 4x - \cos 5x) dx

=\; \frac{\sin 4x}{4} - \frac{\sin 5x}{5} + C

Hence,

a^{b}=4^{5}

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