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AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Section: Mathematics

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Question : 73 of 160

Marks: +1, -0

\int \frac{dx}{(2\sin x+\sec x)^4}=A(1+\tan x)^{-5}

+B(1+\tan x)^{-6}+C(1+\tan x)^{-7}+k

, then A + B + C =

$-\frac{86}{105}$
$-\frac{1}{105}$
$-\frac{26}{105}$
$-\frac{16}{105}$

Solution:

Solve as follow,

I=\int \frac{dx}{(2\sin x+\sec x)^4}

=\int \frac{\sec^4 x}{(2\tan x+\sec^2 x)^4}dx

=\int \frac{1+\tan^2 x}{(1+\tan x)^8} \sec^2 x \, dx

Let

\tan x = t

So,

\sec^2 x \, dx = dt

So,

I = \frac{1+t^2}{(1+t)^8} dt

= \frac{(1+t)^2-2t}{(1+t)^8} dt

=\int \frac{dt}{(1+t)^6}-2\int \frac{dt}{(1+t)^7}+2\int \frac{dt}{(1+t)^8}

=-\frac{1}{5}(1+\tan x)^{-5}+\frac{1}{3}(1+\tan x)^{-6}

-\frac{2}{7}(1+\tan x)^{-7}+k

So,

A=-\frac{1}{5},\;B=\frac{1}{3}

and

C=-\frac{2}{7}

⇒

A+B+C=-\frac{16}{105}

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