Question

Evaluate $\int {\dfrac{{dx}}{{\sin x + \cos x}}}$
A) $\log \tan \left( {\dfrac{\pi }{8} + \dfrac{x}{2}} \right) + C$
B) $\log \tan \left( {\dfrac{\pi }{8} - \dfrac{x}{2}} \right) + C$
C) $\dfrac{1}{{\sqrt 2 }}\log \tan \left( {\dfrac{\pi }{8} + \dfrac{x}{2}} \right) + C$

Answer 1

We will multiply and divide the whole expression with $\dfrac{1}{{\sqrt 2 }}$ and then term is as $\sin \dfrac{\pi }{4}$ or $\cos \dfrac{\pi }{4}$ as per our requirements and then, we will see a formula in it to combine and our integral will become really easy to be solved.

Complete step-by-step answer:
We have $I = \int {\dfrac{{dx}}{{\sin x + \cos x}}}$
We can rewrite this as $I = \int {\dfrac{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)dx}}{{\sin x\left( {\dfrac{1}{{\sqrt 2 }}} \right) + \cos x\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}}$
We can also rewrite this as: $I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\int {\dfrac{{dx}}{{\sin x\cos \dfrac{\pi }{4} + \cos x\sin \dfrac{\pi }{4}}}}$
We also know that $\sin (x + y) = \sin x\cos y + \cos x\sin y$.
So, we will get:- $I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\int {\dfrac{{dx}}{{\sin \left( {x + \dfrac{\pi }{4}} \right)}}}$
We know that $\sin x = \dfrac{1}{{\cos ecx}}$
$I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\int {\cos ec\left( {x + \dfrac{\pi }{4}} \right)dx}$
Now using the formula: $\int {\cos ecx = \ln \left| {\tan \dfrac{x}{2}} \right| + C}$
Hence, we get: $I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\tan \left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right) + C$

Hence, the correct option is (C).

Note: Let us look at the alternate methods for the same question in brief:-
Method 1:
Let $I = \int {\dfrac{{dx}}{{\sin x + \cos x}}}$
We know that we have the formulas: $\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$ and $\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
Now, putting in $\theta = \dfrac{x}{2}$ in these formulas, we will get:-
$\sin x = \dfrac{{2\tan \left( {\dfrac{x}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}$ and $\cos x = \dfrac{{1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}$.
Putting these in the given expression, we will get:-
$\Rightarrow I = \int {\dfrac{{dx}}{{\sin x + \cos x}}} = \int {\dfrac{{dx}}{{\dfrac{{2\tan \left( {\dfrac{x}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)}} + \dfrac{{1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}}}}$
Simplifying the RHS, we will get the following expression:-
$\Rightarrow I = \int {\dfrac{{1 + {{\tan }^2}\left( {\dfrac{x}{2}} \right)dx}}{{2\tan \left( {\dfrac{x}{2}} \right) + 1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}}$
Now, we also know that $1 + {\tan ^2}\theta = {\sec ^2}\theta$. Putting this, we will get the following expression:-
$\Rightarrow I = \int {\dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{2\tan \left( {\dfrac{x}{2}} \right) + 1 - {{\tan }^2}\left( {\dfrac{x}{2}} \right)}}}$
We can write this as:-
$\Rightarrow I = \int {\dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{2 - \left[ {1 - 2\tan \left( {\dfrac{x}{2}} \right) + {{\tan }^2}\left( {\dfrac{x}{2}} \right)} \right]}}}$
We will now use the formula: ${(a - b)^2} = {a^2} + {b^2} - 2ab$. Thus, we will get:-
$\Rightarrow I = \int {\dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{2 - {{\left( {\tan \left( {\dfrac{x}{2}} \right) - 1} \right)}^2}}}}$ ……..(X)
Now, assume that $t = \tan \dfrac{x}{2}$. So, we will have: $dt = \dfrac{d}{{dx}}\left( {\tan \dfrac{x}{2}} \right) \times \dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right)$.
We know that $\dfrac{d}{{d\theta }}(\tan \theta ) = {\sec ^2}\theta$.
So, we have:- $dt = \dfrac{1}{2}{\sec ^2}\dfrac{x}{2}$
Putting all these in (X), we will get:-
$\Rightarrow I = \int {\dfrac{{{{\sec }^2}\left( {\dfrac{x}{2}} \right)dx}}{{2 - {{\left( {\tan \left( {\dfrac{x}{2}} \right) - 1} \right)}^2}}}} = \int {\dfrac{{2dt}}{{2 - {{\left( {t - 1} \right)}^2}}}}$
We can write this as follows by using the formula: ${a^2} - {b^2} = (a - b)(a + b)$.
$\Rightarrow I = 2\int {\dfrac{{dt}}{{\left( {\sqrt 2 + \left( {t - 1} \right)} \right)\left( {\sqrt 2 - \left( {t - 1} \right)} \right)}}}$
Rearranging the terms to get the following expression:-
$\Rightarrow I = \dfrac{1}{{\sqrt 2 }}\int {\left( {\dfrac{1}{{\sqrt 2 + \left( {t - 1} \right)}}} \right.} + \left. {\dfrac{1}{{\sqrt 2 - \left( {t - 1} \right)}}} \right)dt$ ………….(Y)
We also know that $\int {\dfrac{{dx}}{{x + a}}} = \ln |x + a| + C$
Hence, using this in (Y), we will get:-
\Rightarrow I = \dfrac{1}{{\sqrt 2 }}\left\\{ {\ln |\sqrt 2 + (t - 1)| - \ln |\sqrt 2 - (t - 1)|} \right\\} + C
Now, we can just modify this as per our requirements and if there are no options given, we can leave it here only.
Method 2:
We have $I = \int {\dfrac{{dx}}{{\sin x + \cos x}}}$
We can rewrite this as $I = \int {\dfrac{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)dx}}{{\sin x\left( {\dfrac{1}{{\sqrt 2 }}} \right) + \cos x\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}}$
We can also rewrite this as: $I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\int {\dfrac{{dx}}{{\sin x\sin \dfrac{\pi }{4} + \cos x\cos \dfrac{\pi }{4}}}}$
We also know that $\cos (x - y) = \sin x\sin y + \cos x\cos y$.
So, we will get:- $I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\int {\dfrac{{dx}}{{\cos \left( {x - \dfrac{\pi }{4}} \right)}}}$
Let $u = x - \dfrac{\pi }{4}$
$I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\int {\sec udu}$
Now using the formula: $\int {\sec x = \ln \left| {\sec x + \tan x} \right| + C}$
Hence, we get: $I = \left( {\dfrac{1}{{\sqrt 2 }}} \right)\ln \left| {\sec u + \tan u} \right| + C$
Now, just put in the value of u as we assumed and thus, we get our answer.