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AP EAMCET Engineering 2012 Solved Paper

Section: Mathematics

Question : 111 of 160

Marks: +1, -0

\int \sec^{2} x \csc^{4} x dx

= -\frac{1}{3} \cot^{3} x + k \tan x

-2 \cot x + C

k

is equal to

Solution:

Let

I = \int \sec^{2} x \csc^{4} x dx

= \int \frac{1}{\sin^{4} x \cos^{2} x} dx = \int \frac{\sin^{2} x + \cos^{2} x}{\sin^{4} x \cos^{2} x} dx

(\because 1 = \sin^{2} x + \cos^{2} x)

= \int \frac{dx}{\sin^{2} x \cos^{2} x} + \int \frac{dx}{\sin^{4} x}

= \int \frac{(\sin^{2} x + \cos^{2} x) dx}{\sin^{2} x \cos^{2} x} + \int \csc^{4} x dx

(\because 1 = \sin^{2} x + \cos^{2} x)

= \int (\sec^{2} x + \csc^{2} x) dx

+ \int \csc^{2} x (1 + \cot^{2} x) dx

= \tan x - \cot x + \int \csc^{2} x dx

+ \int \csc^{2} x \cot^{2} x dx

= \tan x - \cot x - \cot x - \frac{\cot^{3} x}{3} + C

= -\frac{1}{3} \cot^{3} x + \tan x - 2 \cot x + C

But it given that,

I = -\frac{1}{3} \cot^{3} x + k \tan x - 2 \cot x + C

∴

k = 1

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