Question

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

Answer 1

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

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