Question

$\int\limits_0^{\pi /4} {\dfrac{{\left( {\sin x + \cos x} \right)}}{{9 + 16\sin 2x}}dx}$

Answer 1

First of all we will assume $\sin x - \cos x = t$ and differentiate it with respect to $x$ and from there we will get $dx$. Now for the $\sin 2x$ , we will square the value of $x$ and substitute the value in the original equation and by using the form of integral we will solve it.

Formula used:
Integral form,
$\int {\dfrac{{dx}}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c}$
Log properties used are,
$n\log m = \log {m^n}$
$\log 1 = 0$
$2\sin x\cos x = \sin 2x$

Complete step by step solution:
We have the equation given as $\int\limits_0^{\pi /4} {\dfrac{{\left( {\sin x + \cos x} \right)}}{{9 + 16\sin 2x}}dx}$
Let us assume that $\sin x - \cos x = t$
So now on differentiating the equation with respect to $x$ , we get
$\Rightarrow \cos x + \sin x = \dfrac{{dt}}{{dx}}$
And from here,
$\Rightarrow dx = \dfrac{{dt}}{{\sin x + \cos x}}$
Since, $\sin x - \cos x = t$
Now on squaring both the sides, we get
$\Rightarrow {\left( {\sin x - \cos x} \right)^2} = {t^2}$
On expanding the above equation, we get
$\Rightarrow {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x = {t^2}$
And on solving the equation, we get
$\Rightarrow 1 - 2\sin x\cos x = {t^2}$
And by using the trigonometric formula, we get
$\Rightarrow 1 - \sin 2x = {t^2}$
And from here, $1 - {t^2} = \sin 2x$
So we have the equation,
$\Rightarrow \int\limits_0^{\pi /4} {\dfrac{{\left( {\sin x + \cos x} \right)}}{{9 + 16\sin 2x}}dx}$
Now on putting the values of $dx$ and $\sin 2x$ , we get
$\Rightarrow \int\limits_{ - 1}^0 {\dfrac{{\sin x + \cos x}}{{9 + 16\sin 2x}} \times \dfrac{{dt}}{{\sin x + \cos x}}}$
On canceling the like terms, we get
$\Rightarrow \int\limits_{ - 1}^0 {\dfrac{1}{{9 + 16\left( {1 - {t^2}} \right)}} \cdot dt}$
So on solving the braces, we get
$\Rightarrow \int\limits_{ - 1}^0 {\dfrac{1}{{9 + 16 - 16{t^2}}} \cdot dt}$
So the equation will be as
$\Rightarrow \int\limits_{ - 1}^0 {\dfrac{1}{{25 - 16{t^2}}} \cdot dt}$
Now taking $\dfrac{1}{{16}}$ common, we get
$\Rightarrow \dfrac{1}{{16}}\int\limits_{ - 1}^0 {\dfrac{1}{{\dfrac{{25}}{{16}} - {t^2}}} \cdot dt}$
So the above equation can be written as
$\Rightarrow \dfrac{1}{{16}}\int\limits_{ - 1}^0 {\dfrac{1}{{{{\left( {\dfrac{5}{4}} \right)}^2} - {t^2}}} \cdot dt}$
Since the above equation is of the form $\int {\dfrac{{dx}}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c}$ , and also replacing the $x$ by $t$ and $a$ by $\dfrac{5}{4}$ , we get
$\Rightarrow \dfrac{1}{{16}}\left[ {\dfrac{1}{{2 \cdot \dfrac{5}{4}}}\log \left| {\dfrac{{\dfrac{5}{4} + t}}{{\dfrac{5}{4} - t}}} \right|} \right]_{ - 1}^0$
And on solving it further, we get
$\Rightarrow \dfrac{1}{4}\left[ {\dfrac{1}{{10}}\log \left| {\dfrac{{5 + 4t}}{{5 - 4t}}} \right|} \right]_{ - 1}^0$
Taking the constant term outside, we get
$\Rightarrow \dfrac{1}{{40}}\left[ {\log \left| {\dfrac{{5 + 4t}}{{5 - 4t}}} \right|} \right]_{ - 1}^0$
And on solving the equation further by substituting the values of the limit, we get
$\Rightarrow \dfrac{1}{{40}}\left[ {\log \left| {\dfrac{{5 + 4\left( 0 \right)}}{{5 - 4\left( 0 \right)}}} \right| - \log \left| {\dfrac{{5 + 4\left( { - 1} \right)}}{{5 - 4\left( { - 1} \right)}}} \right|} \right]$
And on solving the multiplication, we get
$\Rightarrow \dfrac{1}{{40}}\left[ {\log \left| {\dfrac{{5 + 0}}{{5 - 0}}} \right| - \log \left| {\dfrac{{5 - 4}}{{5 + 4}}} \right|} \right]$
And on solving the addition, we get
$\Rightarrow \dfrac{1}{{40}}\left[ {\log \dfrac{5}{5} - \log \dfrac{1}{9}} \right]$
So by using the properties of logarithmic, it will be equal to
$\Rightarrow \dfrac{1}{{40}}\left[ {\log 1 - \log {{\left( {\dfrac{1}{9}} \right)}^{ - 1}}} \right]$
And on solving it, we get
$\Rightarrow \dfrac{1}{{40}}\log 9$
**
Hence, the integral $\int\limits_0^{\pi /4} {\dfrac{{\left( {\sin x + \cos x} \right)}}{{9 + 16\sin 2x}}dx}$ will be equal to $\dfrac{1}{{40}}\log 9$.**

Note:
Here we can see that the calculation was very easy. We just need to know the properties we are going to use in it and solving step by step like this will keep you free from error and in this way you can easily solve such type of problem.