Question

Integrate the function: \[\int {\sin x\log \left( {\cos x} \right)} dx\].

Answer 1

Hint:
Here, we need to find the value of the given integral. We will use a substitution method to simplify the integral. Then, we will use integration by parts to find the value of the integral.

Formula Used:
We will use the formula of integration by parts, the integral of the product of two differentiable functions of \[x\] can be written as \[\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx} \], where \[u\] and \[v\] are the differentiable functions of \[x\].

Complete step by step solution:
We will use a substitution method to solve the given integral.
Rewriting the expression \[\int {\sin x\log \left( {\cos x} \right)} dx\], we get
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = \int {\left( 1 \right)\sin x\log \left( {\cos x} \right)} dx\]
Rewriting 1 as the product of \[ - 1\] and \[ - 1\], we get
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = \int {\left( { - 1 \times - 1} \right)\sin x\log \left( {\cos x} \right)} dx\]
We know that we can take constants outside the integral, because \[\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx\].
Therefore, the equation becomes
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = \left( { - 1} \right)\int {\left( { - 1} \right)\sin x\log \left( {\cos x} \right)} dx\]
Simplifying the expression, we get
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - \int {\left( { - \sin x} \right)\log \left( {\cos x} \right)} dx\]
Now, we will use substitution to integrate the expression.
Let \[t = \cos x\].
The derivative of \[\cos x\] with respect to \[x\] is \[ - \sin x\].
Differentiating both sides of the equation \[t = \cos x\] with respect to \[x\], we get
\[ \Rightarrow \dfrac{{dt}}{{dx}} = - \sin x\]
Multiplying both sides of the equation by , we get
\[ \Rightarrow dt = \left( { - \sin x} \right)dx\]
Substituting \[\cos x = t\] and \[\left( { - \sin x} \right)dx = dt\] in the equation \[\int {\sin x\log \left( {\cos x} \right)} dx = - \int {\left( { - \sin x} \right)\log \left( {\cos x} \right)} dx\], we get
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - \int {\log \left( t \right)} dt\]
Thus, we have simplified the expression within the integral.
Now, we will integrate the simplified function using integration by parts.
Rewriting the equation, we get
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - \int {1 \times \log \left( t \right)} dt\]
Using integration by parts, the integral of the product of two differentiable functions of \[x\] can be written as \[\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx} \], where \[u\] and \[v\] are the differentiable functions of \[x\].
Let \[u\] be \[\log t\] and \[v\] be \[1\].
Therefore, by integrating \[\int {1 \times \log \left( t \right)} dt\] by parts, we get
\[ \Rightarrow \int {1 \times \log \left( t \right)} dt = \log t\int {\left( 1 \right)} dt - \int {\left( {\dfrac{{d\left( {\log t} \right)}}{{dt}} \times \int {\left( 1 \right)} dt} \right)} dt\]
We know that the derivative of the function \[\log x\] is \[\dfrac{1}{x}\].
Also, we know that the integral of a constant \[\int {\left( 1 \right)} dx\] is \[x\].
Therefore, we can simplify the integral as
\[ \Rightarrow \int {1 \times \log \left( t \right)} dt = \log t \times t - \int {\left( {\dfrac{1}{t} \times t} \right)} dt\]
Simplifying the expression, we get
\[ \Rightarrow \int {\log \left( t \right)} dt = t\log t - \int {\left( 1 \right)} dt\]
Integrating the expression, we get
\[ \Rightarrow \int {\log \left( t \right)} dt = t\log t - t + K\], where \[K\] is a constant of integration
Substitute \[\int {\log \left( t \right)} dt = t\log t - t + K\] in the equation \[\int {\sin x\log \left( {\cos x} \right)} dx = - \int {\log \left( t \right)} dt\], we get
\[\begin{array}{l} \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - \left( {t\log t - t + K} \right)\\ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - 1\left( {t\log t - t + K} \right)\end{array}\]
Multiplying the terms using the distributive law of multiplication, we get
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - t\log t + t - K\]
Substituting \[t = \cos x\] in the equation, we get
\[\begin{array}{l} \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - \cos x\log \left( {\cos x} \right) + \left( {\cos x} \right) - K\\ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = \left( {\cos x} \right) - \cos x\log \left( {\cos x} \right) - K\\ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = \cos x\left[ {1 - \log \left( {\cos x} \right)} \right] - K\end{array}\]
Substituting \[ - K = C\], we get
\[ \Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = \cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C\]

Thus, we get the value of the integral \[\int {\sin x\log \left( {\cos x} \right)} dx\] as \[\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C\].

Note:
We can verify our answer by differentiating the equation \[\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C\] with respect to \[x\], and checking if the derivative is \[\sin x\log \left( {\cos x} \right)\].
Rewriting the function \[\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C\], we get
\[ \Rightarrow \cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C = \cos x - \cos x\log \left( {\cos x} \right) + C\]
Differentiating the equation with respect to \[x\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{d\left[ {\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C} \right]}}{{dx}} = - \sin x - \dfrac{{d\left[ {\cos x\log \left( {\cos x} \right)} \right]}}{{dx}} + 0\\ \Rightarrow \dfrac{{d\left[ {\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C} \right]}}{{dx}} = - \sin x - \dfrac{{d\left[ {\cos x\log \left( {\cos x} \right)} \right]}}{{dx}}\end{array}\]
Differentiating \[\cos x\log \left( {\cos x} \right)\] using the product rule of differentiation, we get
\[ \Rightarrow \dfrac{{d\left[ {\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C} \right]}}{{dx}} = - \sin x - \left[ {\cos x\dfrac{{d\left\{ {\log \left( {\cos x} \right)} \right\}}}{{dx}} + \dfrac{{d\left\{ {\cos x} \right\}}}{{dx}}\log \left( {\cos x} \right)} \right]\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{{d\left[ {\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C} \right]}}{{dx}} = - \sin x - \cos x\dfrac{{d\left\{ {\log \left( {\cos x} \right)} \right\}}}{{dx}} - \dfrac{{d\left\{ {\cos x} \right\}}}{{dx}}\log \left( {\cos x} \right)\\ \Rightarrow \dfrac{{d\left[ {\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C} \right]}}{{dx}} = - \sin x - \cos x\left[ {\dfrac{1}{{\cos x}} \times \left( { - \sin x} \right)} \right] - \left( { - \sin x} \right)\log \left( {\cos x} \right)\end{array}\]
Multiplying the terms and simplifying the expression, we get
\[ \Rightarrow \dfrac{{d\left[ {\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C} \right]}}{{dx}} = - \sin x + \sin x + \sin x\log \left( {\cos x} \right)\]
Subtracting the terms, we get
\[ \Rightarrow \dfrac{{d\left[ {\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C} \right]}}{{dx}} = \sin x\log \left( {\cos x} \right)\]
Since the derivative of \[\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C\] is \[\sin x\log \left( {\cos x} \right)\], we have verified that the integral of \[\sin x\log \left( {\cos x} \right)\] is \[\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C\].