Find the value of a and b such that$\int{\dfrac{dx}{1\ +\ \sin x}}\ =\ \tan (x+\ a)\ +\ b$.

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Hint: $\dfrac{1}{1+\sin x}$cannot be integrated directly so convert the function such that we can integrate it. Rationalize the given function $\dfrac{1}{1+\sin x}$ before integrating it.

Consider the expression,
$\int{\dfrac{dx}{1\ +\ \sin x}}\ =\tan (x+a)+b$…(1.1)
$\int{\dfrac{dx}{1\ +\ \sin x}}=\tan (x+a)+b$
Multiply $(1-\sin x)$ with both numerator and denominator in L.H.S., we get
$\int{\dfrac{dx}{1\ +\ \sin x}}\times \dfrac{(1\ -\ \sin x)}{(1\ -\ \sin x)}=\tan (x+a)+b$
We know ${{a}^{2\ }}-\ {{b}^{2}}=\ (a\ +\ b)(a\ -\ b)$ so in denominator, we use this formula and we
$\int{\dfrac{(1\ -\ \sin x\ )dx}{{{1}^{2}}\ -{{\sin }^{2}}x}}=\tan (x+a)+b$
We know${{1}^{2}}\ -{{\sin }^{2}}x={{\cos }^{2}}x\ $, so the above equation becomes
$\int{\dfrac{(1\ -\ \sin x\ )dx}{{{\cos }^{2}}x\ }}=\tan (x+a)+b$
Separating the denominator, we get
\[\int (\dfrac{1}{{{\cos }^{2}}x}\ -\ \dfrac{\sin x}{{{\cos }^{2}}x}\ )dx=\tan (x+a)+b\]
We know that $\dfrac{1}{{{\cos }^{2}}x}\ =\ {{\sec }^{2}}x$ , $\dfrac{\sin x}{\cos x}\ =\ \tan x$ and
$\dfrac{\ 1}{\cos x}\ =\ \sec x$, so the above equation becomes
$\int{{{\sec }^{2}}x\ dx\ \ -\ \int{\dfrac{\sin x}{\cos x}\times \dfrac{1}{\cos x}\ dx}}=\tan (x+a)+b$
$\int{{{\sec }^{2}}x\ dx\ \ -\ \int{\tan x\ \sec xdx}}=\tan (x+a)+b$
We know, $\int{{{\sec }^{2}}x\ dx\ =\ \tan x\ }$and$\int{\tan x\ \sec xdx\ =\ \sec x}$, so above
equation becomes

$\Rightarrow \tan x-\sec x+C=\tan (x+a)+b$
Hence, comparing both side we get the value of a & b, so
\[a=0\]; \[b=-\sec x+C\]
Note: In expression$\int{\dfrac{dx}{1\ +\ \sin x}}\ \ $, it’s important to rationalize so that we can get a function after integration which resembles the R.H.S. Without rationalizing, solving the expression becomes complicated and time consuming.
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