Question

Find the value of $\int {\dfrac{{\sec x}}{{\sqrt {\sin \left( {2x + \alpha } \right) + \sin \alpha } }}} dx \\\$
A . $\sqrt {2\sec \alpha \left( {\tan x + \tan \alpha } \right)} + c\\\$
B. $\sqrt {2\sec \alpha \left( {\tan x - \tan \alpha } \right)} + c\\\$
C. $\sqrt {2\sec \alpha \left( {\tan \alpha - \tan x} \right)} + c\\\$
D. $\sqrt {2\sec \alpha \left( {\sec x - \sec \alpha } \right)} + c\\\$

Answer 1

The given function is indefinite since there is no limit given. The indefinite integral of a function f is a differentiable function F whose derivative is equal to the original function f. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals.

Complete step by step solution:
Let the given integral be $I$ such that:
$I = \int {\dfrac{{\sec x}}{{\sqrt {\sin \left( {2x + \alpha } \right) + \sin \alpha } }}} dx - - (i)$
We know the trigonometric addition and subtraction identities of$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, hence by using this identity we can write equation (i) as

$$I = \int {\dfrac{{\sec x}}{{\sqrt {\sin 2x\cos \alpha + \cos 2x\sin \alpha + \sin \alpha } }}} dx \\\ = \int {\dfrac{{\sec x}}{{\sqrt {\sin 2x\cos \alpha + \sin \alpha \left( {\cos 2x + 1} \right)} }}} dx \\\$$

Since, $1 + \cos 2\theta = 2{\cos ^2}\theta$ and also $\sin 2\theta = 2\sin \theta \cos \theta$ hence by using these identities the equation can be further written as:
$I = \int {\dfrac{{\sec x}}{{\sqrt {2\sin x\cos x\cos \alpha + \sin \alpha \left( {2{{\cos }^2}x} \right)} }}} dx$
Substitute $\sin x = \tan x\cos x$ in the above equation, we get

$$I = \int {\dfrac{{\sec x}}{{\sqrt {2\left( {\tan x\cos x} \right)\cos x\cos \alpha + \sin \alpha \left( {2{{\cos }^2}x} \right)} }}} dx \\\ I = \int {\dfrac{{\sec x}}{{\sqrt {2\tan x{{\cos }^2}x\cos \alpha + \sin \alpha \left( {2{{\cos }^2}x} \right)} }}} dx \\\$$

Now take $2{\cos ^2}x$ as common in the denominator, so we get
$I = \dfrac{1}{{\sqrt 2 }}\int {\dfrac{{\sec x}}{{\cos x\sqrt {\tan x\cos \alpha + \sin \alpha } }}} dx{\text{ }}\left[ {\because \tan x = \dfrac{{\sin x}}{{\cos x}}} \right]$
This can be further written as
$I = \dfrac{1}{{\sqrt 2 }}\int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\tan x\cos \alpha + \sin \alpha } }}} dx - - (ii){\text{ }}\left[ {\because \sec x = \dfrac{1}{{\cos x}}} \right]$
Now let $\tan x\cos \alpha + \sin \alpha = {t^2} - - (iii)$
Now differentiate equation (iii) with respect to ‘t’, so we will get

$${\sec ^2}x\cos \alpha dx + 0 = 2tdt \\\ {\sec ^2}x = \dfrac{{2t}}{{\cos \alpha }}dt - - (iv) \\\$$

Now substitute the value of (iii) and (iv) in equation (ii), we get

$$I = \dfrac{1}{{\sqrt 2 }}\int {\dfrac{{\left( {\dfrac{{2t}}{{\cos \alpha }}} \right)}}{{\sqrt {{t^2}} }}} dt \\\ = \dfrac{1}{{\sqrt 2 }}\int {\dfrac{2}{{\cos \alpha }}} dt \\\ = \dfrac{2}{{\sqrt 2 \cos \alpha }}\int {dt} \\\$$

Now by integration, we get
$I = \dfrac{2}{{\sqrt 2 \cos \alpha }}t + c$
Now substitute the value of t in obtained equation from equation (iii), so we get

$$I = \dfrac{{\sqrt 2 }}{{\cos \alpha }}\sqrt {\tan x\cos \alpha + \sin \alpha } + c \\\ = \sqrt {2{{\sec }^2}\alpha \left( {\tan x\cos \alpha + \sin \alpha } \right)} + c \\\ = \sqrt {2\sec \alpha \left( {\tan x\cos \alpha \sec \alpha + \sin \alpha \sec \alpha } \right)} + c \\\ = \sqrt {2\sec \alpha \left( {\tan x + \tan \alpha } \right)} + c \\\$$

Therefore we can say
$\int {\dfrac{{\sec x}}{{\sqrt {\sin \left( {2x + \alpha } \right) + \sin \alpha } }}} dx = \sqrt {2\sec \alpha \left( {\tan x + \tan \alpha } \right)} + c$

Option A is correct.

Important equations used:
$\cos 2\theta = 2{\cos ^2}\theta - 1$
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\int {dx = x + c}$
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\sec x = \dfrac{1}{{\cos x}}$

Note:
While substituting the real parameter of the question with the auxiliary parameter, one should be sure that it will not make the problem more complex. However, selecting an auxiliary parameter completely depends on the individual point of view.