Question

Evaluate the following integral
$\int {\dfrac{{\sin 3x}}{{\cos 4x\cos x}}dx}$ is equal to
$\left( a \right) - \dfrac{1}{4}\log \left| {\cos 4x} \right| - \log \left| {\cos x} \right| + c$
$\left( b \right) - \dfrac{1}{4}\log \left| {\cos 4x} \right| + \log \left| {\cos x} \right| + c$
$\left( c \right) - \dfrac{1}{4}\log \left| {\sin 4x} \right| + \log \left| {\cos x} \right| + c$
$\left( c \right)\dfrac{1}{4}\log \left| {\sin 4x} \right| - \log \left| {\cos x} \right| + c$

Answer 1

In this particular question write sin 3x = sin (4x - x), then apply the basic sine rule i.e. sin (A – B) = sin A cos B – cos A sin B, then use the property that, tan x = (sin x/cos x), so use these properties to reach the solution of the question.

Complete step-by-step answer:
Given integral
$\int {\dfrac{{\sin 3x}}{{\cos 4x\cos x}}dx}$
Let, $I = \int {\dfrac{{\sin 3x}}{{\cos 4x\cos x}}dx}$
Now sin 3x is written as, sin (4x - x) so we have,
$\Rightarrow I = \int {\dfrac{{\sin \left( {4x - x} \right)}}{{\cos 4x\cos x}}dx}$
Now as we know that sin (A – B) = sin A cos B – cos A sin B so use this property in the above integral we have,
$\Rightarrow I = \int {\dfrac{{\sin 4x\cos x - \cos 4x\sin x}}{{\cos 4x\cos x}}dx}$
Now separate the integral and cancel out the common terms from the numerator and denominator we have,
$\Rightarrow I = \int {\dfrac{{\sin 4x}}{{\cos 4x}}dx} - \int {\dfrac{{\sin x}}{{\cos x}}dx}$
Now as we know that tan x = (sin x/cos x) so use this property in the above integral we have,
$\Rightarrow I = \int {\tan 4xdx} - \int {\tan xdx}$
Now as we know that $\int {\tan ax} dx = - \dfrac{{\log \left| {\cos ax} \right|}}{a} + c$ where c is some arbitrary integration constant, so use this property in the above integral we have,
$\Rightarrow I = - \dfrac{1}{4}\log \left| {\cos 4x} \right| - \left( { - \dfrac{1}{1}\log \left| {\cos x} \right|} \right) + c$
$\Rightarrow I = - \dfrac{1}{4}\log \left| {\cos 4x} \right| + \log \left| {\cos x} \right| + c$
So this is the required integral.

So, the correct answer is “Option b”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the integration of tan ax which is given as $\int {\tan ax} dx = - \dfrac{{\log \left| {\cos ax} \right|}}{a} + c$ where c is some arbitrary integration constant, so simply apply this as above we will get the required answer.