Question

Evaluate $\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x + \cos x}}{{16 + 9\sin 2x}}dx}$

Answer 1

Hint : In order to determine the answer of above definite integral use the method of Integration by substitution by substituting $\sin x - \cos x$ with $t$ . With the use of this assumption, rewrite the whole integral in terms of $t$ along with the limits. Use the standard formula of integration $\int {\dfrac{1}{{{a^2} - {{\left( {bx} \right)}^2}}} = \dfrac{1}{{2ab}}} \log \left| {\dfrac{{a + bx}}{{a - bx}}} \right|$ to evaluate the integral and then put the limits as [Upper limit – lower limit] to obtain the final answer.
Formula:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$

Complete step-by-step answer :
We are Given integral $\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x + \cos x}}{{16 + 9\sin 2x}}dx}$ -(1)
Here we are using Integration by substitution method to solve the above integral
Now, let’s assume $\sin x - \cos x = t$ -(2)
Calculating the first derivative of the above assumed equation we get,
$\left( {\cos x + \sin x} \right)dx = dt$
As per the assumption, we have
${\left( {\sin x - \cos x} \right)^2} = {t^2}$
With the help of identity ${\left( {A - B} \right)^2} = {A^2} + {B^2} + 2AB$ and trigonometric identity $2\sin x\cos x = \sin 2x$ ,we can rewrite the expression as
$\sin 2x = 1 - {t^2}$
Since we have assumed some $x$ expression in terms of $t$ , so the limits of integrations will also change accordingly. So
$x = 0,t = - 1$ as $\sin \left( 0 \right) - \cos \left( { - 1} \right) = t \Rightarrow t = - 1$
Similarly
$x = \dfrac{\pi }{4},t = 0$ as $\sin \left( {\dfrac{\pi }{4}} \right) - \cos \left( {\dfrac{\pi }{4}} \right) = t \Rightarrow t = 0$
Now with the help of assumptions and results obtained above and along with the proper limits we can rewrite our original integral expression as
$\therefore I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x + \cos x}}{{16 + 9\sin 2x}}dx} \\\ I = \int\limits_{ - 1}^0 {\dfrac{1}{{16 + 9\left( {1 - {t^2}} \right)}}dt} \\\$
Simplifying further ,
$I = \int\limits_{ - 1}^0 {\dfrac{1}{{25 - 9{t^2}}}dt}$
Rewriting the above integral , we get
$I = \int\limits_{ - 1}^0 {\dfrac{1}{{{{\left( 5 \right)}^2} - {{\left( {3t} \right)}^2}}}dt}$
Now using the rule of integration $\int {\dfrac{1}{{{a^2} - {{\left( {bx} \right)}^2}}} = \dfrac{1}{{2ab}}} \log \left| {\dfrac{{a + bx}}{{a - bx}}} \right|$ , we obtain the integration as
$I = \left[ {\dfrac{1}{{30}}\log \left| {\dfrac{{5 + 3t}}{{5 - 3t}}} \right|} \right]_{ - 1}^0$
Assigning limits to the above integral
Limits are calculated as [Upper limit – lower limit]
$I = \dfrac{1}{{30}}\left[ {\log \left| {\dfrac{{5 + 3\left( 0 \right)}}{{5 - 3\left( 0 \right)}}} \right| - \log \left| {\dfrac{{5 + 3\left( { - 1} \right)}}{{5 - 3\left( { - 1} \right)}}} \right|} \right] \\\ = \dfrac{1}{{30}}\left[ {\log \left| {\dfrac{5}{5}} \right| - \log \left| {\dfrac{{5 - 3}}{{5 + 3}}} \right|} \right] \\\ = \dfrac{1}{{30}}\left[ {\log \left| 1 \right| - \log \left| {\dfrac{2}{8}} \right|} \right] \\\$
As we know the value for $\log \left( 1 \right) = 0$
$= \dfrac{1}{{30}}\left[ {0 - \log \left| {\dfrac{1}{4}} \right|} \right] \\\ = - \dfrac{1}{{30}}\log \dfrac{1}{4} \;$
$\therefore I = - \dfrac{1}{{30}}\log \dfrac{1}{4}$
Therefore, the value of the given integral $\int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x + \cos x}}{{16 + 9\sin 2x}}dx}$ is equal to $- \dfrac{1}{{30}}\log \dfrac{1}{4}$
So, the correct answer is “ $- \dfrac{1}{{30}}\log \dfrac{1}{4}$ ”.

Note : 1.Different types of methods of Integration:
I.Integration by Substitution
II.Integration by parts
III.Integration of rational algebraic function by using partial fraction
2. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.
If $\varphi (x)$ is a continuously differentiable function, then to evaluate integrals of the form.
$\int {f(\varphi (x))\,{\varphi ^1}(x)dx}$, we substitute $\varphi (x)$ =t and ${\varphi ^1}(x)dx = dt$
This substitution reduces the above integral to $\int {f(t)\,dt}$. After evaluating this integral we substitute back the value of t.