Question

Find the integral of $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}}} $

Answer 1

Hint: In this question we can see that we have trigonometric ratios as sine and cosine are trigonometric functions. We will use some of the basic formulas to solve this question. We will use the formula: $\int\limits_b^a {f(x)dx = } \int\limits_b^a {(a + b - x)dx} $, where $a,b$ are the limits of the integration.

Complete step by step answer:
Let us assume that
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}}} dx$
We will assume this as equation (one).
Now we will use this formula i.e.
$\int\limits_b^a {f(x)dx = } \int\limits_b^a {(a + b - x)dx} $ ,
By comparing this with the above question we have
$a = \dfrac{\pi }{2},b = 0$
So we can write $x = \left( {\dfrac{\pi }{2} + 0 - x} \right)$

We will substitute this value in the question and it can be written as:
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin \left( {\dfrac{\pi }{2} - x} \right) - \cos \left( {\dfrac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\dfrac{\pi }{2} - x} \right)\cos \left( {\dfrac{\pi }{2} - x} \right)}}} dx$
Now we know the identity that
$\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x$
And, $\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x$
So by applying this in the above question we can write :
$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x - \sin x}}{{1 + \cos x\sin x}}} dx$
This is our second equation.
We will now add both our equation, so we can write:
\[I + I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}}} dx + \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x - \sin x}}{{1 + \cos x\sin x}}} dx\]

Since the denominator of both the terms in the RHS, so we will take the LCM and add them:
$2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x - \cos x + \cos x - \sin x}}{{1 + \cos x\sin x}}} dx$
On simplifying it gives us:
$2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{0}{{1 + \cos x\sin x}}} dx$
Now we know that any fraction in which the numerator is zero, it gives the value as zero i.e.
$\dfrac{0}{a} = 0$
So we can write;
$2I = 0$
By isolating the term in LHS, we have
$I = \dfrac{0}{2} = 0$

Hence the value of the above given integral in the question is $0$.

Note:We should note that in the above solution we have written
\[\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x\] . We can solve this with the addition formula i.e. $\sin (A - B) = \sin A\cos B - \cos A\sin B$
So by applying this we can write:
$\sin \left( {\dfrac{\pi }{2} - x} \right) = \sin \left( {\dfrac{\pi }{2}} \right)\cos x - \cos \left( {\dfrac{\pi }{2}} \right)\sin x$
Now we know that here
$\dfrac{\pi }{2} = {90^ \circ }$ and so we have the trigonometric value i.e.
$\sin {90^ \circ } = 1$
And,
$\cos {90^ \circ } = 0$
So by putting this values in the formula we have:
$1 \times \cos x - 0 = \cos x$
Hence we get:
$\sin \dfrac{\pi }{2} - x = \cos x$
We can write this directly as
\[\sin \left( {\dfrac{\pi }{2} - x} \right) = \sin ({90^ \circ } - x) = \cos x\]
We can write directly the same for cosine too.