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

Last updated date: 11th Sep 2024
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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 $\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C$.