\(∫\dfrac{1}{cosx(sinx+2cosx)} dx=\)
\(ln|(1-tanx)|+C \)
\(ln |3+sinx|+C\)
\(ln |2+tanx|+C\)
\(ln |1+2secx|+C\)
\(ln |2-tanx|+C\)
The Correct Option is C
Approach Solution - 1
Given: \(∫\dfrac{1}{cosx(sinx+2cosx)} dx\)
take, \(I = ∫\dfrac{1}{cosx(sinx+2cosx)} dx\)
\(= ∫\dfrac{1}{cosx.sinx+2cos^{2}x)} dx\)
\(= ∫\dfrac{sec^{2}x}{tanx+2} dx\) ⇢⇢[by dividing \(cos^{2}x\) on both numerator and denominator]
take \(tanx +2 = u\), then derivate both sides w.r.t x we get
\(sec^{2}x dx= du\)
Bye applying this we can write , \(∫\dfrac{1}{u} du\)
\(= ln|u|\)
\(= ln|(tanx+2)|+C\)
\(= ln|(2+tanx)|+C\)
Approach Solution -2
Let the integral be
\[ I = \int \frac{dx}{\cos x (\sin x + 2 \cos x)} \]
\[ I = \int \frac{\sec x \, dx}{2 + \tan x} \]
Let \( u = 2 + \tan x \). Then \( du = \sec^2 x \, dx \).
We have \( \sec x = \sqrt{1 + \tan^2 x} = \sqrt{1 + (u-2)^2} \).
\[ I = \int \frac{\sqrt{1 + (u-2)^2}}{u} \frac{du}{\sqrt{1+(u-2)^2}} = \int \frac{du}{u} = \ln|u| + C = \ln|2 + \tan x| + C \]
Final Answer: The final answer is \( {\ln|2+\tan x|+C} \).
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C