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

Integrate the function: ∫sin⁡xlog⁡(cos⁡x)dx\int {\sin x\log \left( {\cos x} \right)} dx.

Explanation

Solution

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 xx can be written as ∫uvdx=u∫vdx−∫(d(u)dx×∫vdx)dx\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx} , where uu and vv are the differentiable functions of xx.

Complete step by step solution:
We will use a substitution method to solve the given integral.
Rewriting the expression ∫sin⁡xlog⁡(cos⁡x)dx\int {\sin x\log \left( {\cos x} \right)} dx, we get
⇒∫sin⁡xlog⁡(cos⁡x)dx=∫(1)sin⁡xlog⁡(cos⁡x)dx\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 - 1 and −1 - 1, we get
⇒∫sin⁡xlog⁡(cos⁡x)dx=∫(−1×−1)sin⁡xlog⁡(cos⁡x)dx\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 ∫af(x)dx=a∫f(x)dx\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx.
Therefore, the equation becomes
⇒∫sin⁡xlog⁡(cos⁡x)dx=(−1)∫(−1)sin⁡xlog⁡(cos⁡x)dx\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
⇒∫sin⁡xlog⁡(cos⁡x)dx=−∫(−sin⁡x)log⁡(cos⁡x)dx\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⁡xt = \cos x.
The derivative of cos⁡x\cos x with respect to xx is −sin⁡x - \sin x.
Differentiating both sides of the equation t=cos⁡xt = \cos x with respect to xx, we get
⇒dtdx=−sin⁡x\Rightarrow \dfrac{{dt}}{{dx}} = - \sin x
Multiplying both sides of the equation by , we get
⇒dt=(−sin⁡x)dx\Rightarrow dt = \left( { - \sin x} \right)dx
Substituting cos⁡x=t\cos x = t and (−sin⁡x)dx=dt\left( { - \sin x} \right)dx = dt in the equation ∫sin⁡xlog⁡(cos⁡x)dx=−∫(−sin⁡x)log⁡(cos⁡x)dx\int {\sin x\log \left( {\cos x} \right)} dx = - \int {\left( { - \sin x} \right)\log \left( {\cos x} \right)} dx, we get
⇒∫sin⁡xlog⁡(cos⁡x)dx=−∫log⁡(t)dt\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
⇒∫sin⁡xlog⁡(cos⁡x)dx=−∫1×log⁡(t)dt\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 xx can be written as ∫uvdx=u∫vdx−∫(d(u)dx×∫vdx)dx\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx} , where uu and vv are the differentiable functions of xx.
Let uu be log⁡t\log t and vv be 11.
Therefore, by integrating ∫1×log⁡(t)dt\int {1 \times \log \left( t \right)} dt by parts, we get
⇒∫1×log⁡(t)dt=log⁡t∫(1)dt−∫(d(log⁡t)dt×∫(1)dt)dt\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\log x is 1x\dfrac{1}{x}.
Also, we know that the integral of a constant ∫(1)dx\int {\left( 1 \right)} dx is xx.
Therefore, we can simplify the integral as
⇒∫1×log⁡(t)dt=log⁡t×t−∫(1t×t)dt\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
⇒∫log⁡(t)dt=tlog⁡t−∫(1)dt\Rightarrow \int {\log \left( t \right)} dt = t\log t - \int {\left( 1 \right)} dt
Integrating the expression, we get
⇒∫log⁡(t)dt=tlog⁡t−t+K\Rightarrow \int {\log \left( t \right)} dt = t\log t - t + K, where KK is a constant of integration
Substitute ∫log⁡(t)dt=tlog⁡t−t+K\int {\log \left( t \right)} dt = t\log t - t + K in the equation ∫sin⁡xlog⁡(cos⁡x)dx=−∫log⁡(t)dt\int {\sin x\log \left( {\cos x} \right)} dx = - \int {\log \left( t \right)} dt, we get
⇒∫sin⁡xlog⁡(cos⁡x)dx=−(tlog⁡t−t+K) ⇒∫sin⁡xlog⁡(cos⁡x)dx=−1(tlog⁡t−t+K)\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
⇒∫sin⁡xlog⁡(cos⁡x)dx=−tlog⁡t+t−K\Rightarrow \int {\sin x\log \left( {\cos x} \right)} dx = - t\log t + t - K
Substituting t=cos⁡xt = \cos x in the equation, we get
⇒∫sin⁡xlog⁡(cos⁡x)dx=−cos⁡xlog⁡(cos⁡x)+(cos⁡x)−K ⇒∫sin⁡xlog⁡(cos⁡x)dx=(cos⁡x)−cos⁡xlog⁡(cos⁡x)−K ⇒∫sin⁡xlog⁡(cos⁡x)dx=cos⁡x[1−log⁡(cos⁡x)]−K\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 - K = C, we get
⇒∫sin⁡xlog⁡(cos⁡x)dx=cos⁡x[1−log⁡(cos⁡x)]+C\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 ∫sin⁡xlog⁡(cos⁡x)dx\int {\sin x\log \left( {\cos x} \right)} dx as cos⁡x[1−log⁡(cos⁡x)]+C\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C.

Note:
We can verify our answer by differentiating the equation cos⁡x[1−log⁡(cos⁡x)]+C\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C with respect to xx, and checking if the derivative is sin⁡xlog⁡(cos⁡x)\sin x\log \left( {\cos x} \right).
Rewriting the function cos⁡x[1−log⁡(cos⁡x)]+C\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C, we get
⇒cos⁡x[1−log⁡(cos⁡x)]+C=cos⁡x−cos⁡xlog⁡(cos⁡x)+C\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 xx, we get
⇒d[cos⁡x[1−log⁡(cos⁡x)]+C]dx=−sin⁡x−d[cos⁡xlog⁡(cos⁡x)]dx+0 ⇒d[cos⁡x[1−log⁡(cos⁡x)]+C]dx=−sin⁡x−d[cos⁡xlog⁡(cos⁡x)]dx\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⁡xlog⁡(cos⁡x)\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
⇒d[cos⁡x[1−log⁡(cos⁡x)]+C]dx=−sin⁡x+sin⁡x+sin⁡xlog⁡(cos⁡x)\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
⇒d[cos⁡x[1−log⁡(cos⁡x)]+C]dx=sin⁡xlog⁡(cos⁡x)\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[1−log⁡(cos⁡x)]+C\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C is sin⁡xlog⁡(cos⁡x)\sin x\log \left( {\cos x} \right), we have verified that the integral of sin⁡xlog⁡(cos⁡x)\sin x\log \left( {\cos x} \right) is cos⁡x[1−log⁡(cos⁡x)]+C\cos x\left[ {1 - \log \left( {\cos x} \right)} \right] + C.